Complicated Constraint - Simple Inequality
Problem
Solution 1
We know that
$\displaystyle\begin{align} (a+b+c)^3 &= \sum_{cycl}a^3+\prod_{cycl}(a+b)\\ &\ge \sum_{cycl}a^3+\frac{8}{\sqrt[8]{a^3+b^3+c^3}}\\ &\ge (1+8)\sqrt[9]{ \sum_{cycl}a^3\left( \frac{1}{\sqrt[8]{a^3+b^3+c^3}}\right)^8}\\ &=9. \end{align}$
Hence $a+b+c\ge 9^{1/3}.$
The equality is attained only if $a=b=c=\displaystyle\frac{1}{\sqrt[3]{3}}.$
Solution 2
The given condition can be rewrtitten as
(1)
$\displaystyle\sqrt[8]{(a^3+b^3+c^3)(a+b)^8(b+c)^8(c+a)^8\left(\frac{3}{8}\right)^8}\ge 1.$
Applying the AM-GM inequalty, we obtain
$\displaystyle\begin{align} &\sqrt[8]{(a^3+b^3+c^3)\underbrace{\frac{3}{8}(a+b)(b+c)(c+a)\cdots\frac{3}{8}(a+b)(b+c)(c+a)}_{8\,\text{times}}}\\ &\le\frac{1}{9}\left[(a^3+b^3+c^3)+\underbrace{\frac{3}{8}(a+b)(b+c)(c+a)+\ldots+\frac{3}{8}(a+b)(b+c)(c+a)}_{8\,\text{times}}\right]\\ &=\frac{1}{9}\left[(a^3+b^3+c^3)+3(a+b)(b+c)(c+a)\right]\\ &=\frac{1}{9}(a+b+c)^3. \end{align}$
Combining this with (1) we infer that
$(a+b+c)^3\ge 9,$
and the result follows.
The equality occurs if and only if $a=b=c=\displaystyle\frac{1}{\sqrt[3]{3}}.$
Illustration
Acknowledgment
The problem above (from the Romanian Mathematical Magazine, #SP036, posed December 14, 2016) has been kindly posted at the CutTheKnotMath facebook page by Dan Sitaru, along with two solutions. The problem has been proposed by Nguyen Viet Hung. Two practically identical solutions (Solution 1) have been submitted by Anas Adlany (Morocco) and Soumitra Mandal (India). Solution 2 is by Nguyen Viet Hung. The illustration is by Gary Davis.
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- Algebraic-Geometric Inequality $\left(\sqrt{x^2-\sqrt{3}xy+y^2} + \sqrt{y^2-\sqrt{2}yz+z^2} \ge \sqrt{z^2-zx+x^2}\right)$
- One Inequality - Two Domains $\left(\displaystyle 3\prod_{cycl}(a^2+ab+b^2)\ge\left(\sum_{cycl}a\right)^2\cdot\left(\sum_{cycl}ab\right)^2\right)$
- Radicals, Radicals, And More Radicals in an Inequality $\bigg(\displaystyle\gamma=\frac{\sqrt[4]{xz}}{\sqrt{x}+\sqrt{z}}.\,$ Prove that $\sqrt{x}+\sqrt{y}+\sqrt{z}\ge 2\gamma\sqrt{3(x+y+z)}\bigg)$
- An Inequality in Triangle and In General $\left(\displaystyle\sum_{cycl}\frac{\cot A\,\cot^3B}{\cot^2B+2\cot^2A}+2\sum_{cycl}\frac{\cot^2A\cot B}{\cot A+2\cot B}\ge 1\right)$
- Dan Sitaru's Cyclic Inequality In Many Variables $\left(\displaystyle a+b+c+d\le \frac{a^5+b^5+c^5+d^5}{abcd}\right)$
- An Inequality on Circumscribed Quadrilateral $\left(s\ge 4R\right)$
- An Inequality with Fractions $\left(\displaystyle m\le\frac{a_1+a_2+\ldots+a_n}{b_1+b_2+\ldots+b_n}\le M\right)$
- An Inequality with Complex Numbers of Unit Length $\left(|a-b|+|a-c|\ge |a+b|+|a+c|\right)$
- An Inequality with Complex Numbers of Unit Length II $\left(|a^2+bc|\ge |b+c|\right)$
- Le Khanh Sy's Problem $\left(xa^2+yb^2+zc^2\ge 2m\right)$
- An Inequality Not in Triangle $\left(\displaystyle\sqrt{a^2+b^2-ab\sqrt{2}}+\sqrt{b^2+c^2-bc\sqrt{3}}+\sqrt{c^2+d^2-\frac{cd(\sqrt{6}+\sqrt{2})}{2}}\ge\sqrt{a^2+d^2}\right)$
- An Acyclic Inequality in Three Variables $\left(\displaystyle \frac{(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2}{a^2+b^2+c^2+ab+bc+ca}\geq 3(a-b)(b-c)\right)$
- An Inequality with Areas, Norms, and Complex Numbers $\left(\displaystyle \frac{(ad-bc)(3(a^2+b^2)(c^2+d^2)-4(ad-bc)^2)}{\left((a^2+b^2)(c^2+d^2)\right)^{\frac{3}{2}}}\le 1\right)$
- Darij Grinberg's Inequality In Three Variables $\left(a^2+b^2+c^2+2abc+1\ge 2(ab+bc+ca)\right)$
- Small Change Makes Big Difference $\left(\displaystyle\frac{1}{\displaystyle \sqrt{1+a^2-\frac{(a-b)^2}{2}}}+\frac{1}{\displaystyle \sqrt{1+b^2-\frac{(a-b)^2}{2}}}\ge\frac{2}{\sqrt{1+ab}}\right)$
- Inequality with Two Variables? Think Again $\left(\displaystyle\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\le\frac{2}{\sqrt{1+ab}}\right)$
- A Problem From a Mongolian Olympiad for Grade 11 $\left(\displaystyle \frac{a}{3a+2b^3}+ \frac{b}{3b+2c^3}+ \frac{c}{3c+2a^3}\le\frac{1}{5}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\right)$
- Sitaru--Schweitzer Inequality
$\left(\displaystyle \left(\int_{a}^{b}f(x)dx\right)\left(\int_{a}^{b}\frac{1}{f(x)}dx\right)\le\frac{(m+M)^2}{4mM}(b-a)^2\right)$
- Pólya-Szegö Inequality $\left(\displaystyle \frac{\displaystyle \left(\sum_{k=1}^{n}a^2_{k}\right)\left(\sum_{k=1}^{n}b^2_{k}\right)}{\displaystyle\left(\sum_{k=1}^{n}a_{k}b_{k}\right)^2}\le\left(\frac{\displaystyle \sqrt{\frac{M_1M_2}{m_1m_2}}+\sqrt{\frac{m_1m_2}{M_1M_2}}}{2}\right)^2\right)$
- Kantorovich Inequality $\left(\displaystyle \left(\sum_{k=1}^{n}\gamma_ku_k^2\right)\left(\sum_{k=1}^{n}\frac{1}{\gamma_k}u_k^2\right)\le\frac{1}{4}\left(\sqrt{\frac{M}{m}}+\sqrt{\frac{m}{M}}\right)^2\left(\sum_{k=1}^{n}u^2_{k}\right)^2\right)$
- Greub-Rheinboldt Inequality $\left(\displaystyle \left(\sum_{k=1}^{n}a_k^2u_k^2\right)\left(\sum_{k=1}^{n}b_k^2u_k^2\right)\le\frac{(M_1M_2+m_1m_2)^2}{4m_1m_2M_1M_2}\left(\sum_{k=1}^{n}a_kb_ku^2_{k}\right)^2\right)$
- An Inequality with Cyclic Sums And Products $\left(\small{\displaystyle \sum_{cycl}\frac{a^2}{(b+c+d+e)(a-b)(a-c)(a-d)(a-e)}\lt\frac{(a+b+c+d+e)^2}{1024abcde}}\right)$
- Problem 1 From the 2016 Pan-African Math Olympiad $\left(\displaystyle \sum_{cycl}\frac{1}{(x+1)^2+y^2+1}\le\frac{1}{2}\right)$
- An Inequality with Integrals and Radicals $\left(\displaystyle \Bigr(\int_0^1 \sqrt[3]{f(x)}dx\Bigr)\Bigr(\int_0^1 \sqrt[5]{f(x)}dx\Bigr)\Bigr(\int_0^1 \sqrt[7]{f(x)}dx\Bigr)\leq 1\right)$
- Twin Inequalities in Four Variables: Twin 1 $\left(\displaystyle (ac+bd)^2\le\left(b\sqrt[5]{ab^4}+d\sqrt[5]{cd^4}\right)\left(a\sqrt[5]{a^4b}+c\sqrt[5]{c^4d}\right)\right)$
- Twin Inequalities in Four Variables: Twin 2 $\left(\displaystyle (a\sqrt[3]{a^2b}+c\sqrt[3]{c^2d})(b\sqrt[3]{ab^2}+d\sqrt[3]{cd^2})\le (a^2+c^2)(b^2+d^2)\right)$
- Simple Inequality with a Variety of Solutions $\left(\displaystyle \sum_{cycl}\left(\frac{\ln x}{\ln y\ln z}+\frac{\ln y}{\ln z\ln x}\right)\ge\frac{18}{\ln (xyz)}\right)$
- A Partly Cyclic Inequality in Four Variables $\left(\displaystyle \sum_{cycl}xe^x\ge (x+y+2)e^{x+y+2}+(z+t-2)\sqrt[3]{e^{z+t-2}}\right)$
- Dan Sitaru's Inequality by Induction $\left(\displaystyle\begin{align}&\small{\frac{3}{a+1}+\frac{3}{b+1}+\frac{2}{c+1}+\frac{1}{d+1}}\\ &\small{\qquad\le 6+\frac{1}{a+b+1}+\frac{1}{a+b+c+1}+\frac{1}{a+b+c+d+1}}\end{align}\right)$
- An Inequality in Three (Or Is It Two) Variables $\left(\displaystyle \frac{(x+y)^2}{(x\sin^2z+y\cos^2z)(x\cos^2z+y\sin^2z)}+\frac{x}{y}+\frac{y}{x}\geq 6\right)$
- An Inequality in Four Weighted Variables $\left(\displaystyle (a+c)^c(b+d)^d(c+d)^{c+d}\le c^cd^d(a+b+c+d)^{c+d}\right)$
- An Inequality in Fractions with Absolute Values $\left(\displaystyle \omega\lt\frac{1}{3}\left(\sum_{cycl}\frac{a|a|-b|b|}{a-b}\right)\lt 2\Omega\right)$
- Inequalities with Double And Triple Integrals $\left(\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\cos\left(\frac{x+y}{2}\right)dxdy\ge\frac{\pi}{2}\right)$
- An Old Inequality $\left(\displaystyle \frac{4ab}{(a+b)^2}\ge\cos\left(\frac{\pi}{2}\cdot\frac{a-b}{a+b}\right)\right)$
- Dan Sitaru's Amazing, Never Ending Inequality $\left(\displaystyle \small{\sum_{cycl}\left(\frac{a}{b}\right)^2\cdot\sum_{cycl}\left(\frac{a}{b}\right)^4\cdot\sum_{cycl}\left(\frac{a}{b}\right)^8\ge\sum_{cycl}\left(\frac{a}{c}\right)\cdot\sum_{cycl}\left(\frac{b}{a}\right)\cdot\sum_{cycl}\left(\frac{b}{c}\right)}\right)$
- Leo Giugiuc's Exercise $\left(\displaystyle x\sin x+x^2\cos x\le 2\sin^2x\right)$
- Another Inequality with Logarithms, But Not Really $\left(\displaystyle \sum_{cycl}\frac{\log_y^3x+\log_z^3y}{\log_y^2x+\log_zx+\log_z^2y}\geq 2\right)$
- An Inequality Solved by Changing Appearances $\left(\displaystyle \sum_{cycl}a^2\cdot\sum_{cycl}\frac{1}{x^2}+\frac{\displaystyle 2\sum_{cycl}ab\cdot\sum_{cycl}x}{xyz}\ge 0\right)$
- Distances to Three Points on a Circle $\left(3\le |z-a|+|z-b|+|z-c|\le 4\right)$
- An Inequality with Powers And Logarithm $\left(\displaystyle \frac{a}{b}+\frac{a^2}{b^2}+\frac{a^3}{b^3}+12\ln b\ge\frac{b}{a}+\frac{b^2}{a^2}+\frac{b^3}{a^3}+12\ln a\right)$
- Four Integrals in One Inequality $\left(\displaystyle \small{\left(\int_a^bxf(x)dx\right)\left(\int_a^bf^2(x)dx\right)\left(\int_a^bx^3f(x)dx\right)\ge\frac{a^2b^2}{b-a}\left(\int_a^bf(x)dx\right)^4}\right)$
- Same Integral, Three Intervals $\left(\displaystyle\small{I(u,v)=\int_u^v\left(\arctan\left(\frac{u\sin x}{v+u\cos x}\right)+\arctan\left(\frac{v\sin x}{u+v\cos x}\right)\right)dx}\right)$
- Dorin Marghidanu's Inequality with Generalization $\left(\displaystyle (x+y)^2+(y+z)^2+(z+x)^2+12ab\le 4(a+b)(x+y+z)\right)$
- Dan Sitaru's Inequality with Three Related Integrals and Derivatives $\left(\displaystyle\small{\left(\int_0^af(x)dx\right)^4\leq \frac{a^8}{60}\left(\int_0^a \left(f'(x)\right)^2 dx\right)\left(\int_0^a \left(f''(x)\right)^2dx\right)}\right)$
- An Inequality in Two Or More Variables $\left(\displaystyle \frac{a}{1+a}+\frac{b}{(1+a)(1+b)}+\frac{c}{(1+a)(1+b)(1+c)}\geq \frac{7}{8}\right)$
- An Inequality in Two Or More Variables II $\left(\displaystyle (a+1)^{a+1}\cdot (b+1)^{b+1}\cdot (c+1)^{c+1}\le e^{a+b+c}\cdot\sqrt{e^{a^2+b^2+c^2}}\right)$
- A Not Quite Cyclic Inequality $\left(\displaystyle \frac{a^2+b^2+c^2}{a+b+c} \le \frac{ab+bc+ca}{a+b+c} + |a-b|+|b-c|\right)$
- Dan Sitaru's Inequality: From Three Variables to Many in Two Ways $\left(\displaystyle a+b+c\ge\frac{3}{2}\right)$
- An Inequality with Sines But Not in a Triangle $\left(\displaystyle \prod_{cycl}\Bigr(a^2\sin \frac{2\pi}{a}+(a+1)^2\sin \frac{2\pi}{a+1}\Bigr)\gt 2^{16}\right)$
- An Inequality with Angles and Integers $\left(\displaystyle k^2\tan \alpha +l^2\tan \beta \geq \frac{2kl}{\sin (\alpha+\beta)}-(k^2+l^2)\cot (\alpha+\beta)\right)$
- Sladjan Stankovik's Inequality In Four Variables $\left(\displaystyle 2\sum_{cycl}a^2-3\frac{\displaystyle \sum_{cycl}a^3}{\displaystyle \sum_{cycl}a}\le\sum_{all}ab\right)$
- An Inequality with Two Pairs of Triplets $\left(\displaystyle (a^2+b^2+c^2)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\frac{2(ab+bc+ca)(x+y+z)}{xyz}\ge 0\right)$
- A Refinement of Turkevich's Inequality $\left(\displaystyle a^2+b^2+c^2+d^2+\frac{32abcd}{(a+b+c+d)^2}\ge \sum_{sym}ab\right)$
- Dan Sitaru's Exercise with Pi and Ln $\left(\displaystyle \small{\arctan(z-x)+\arctan(z-y)+\arctan(y-x) \lt \frac{\pi}{2} - \ln(2)}\right)$
- Leo Giugiuc's Cyclic Quickie in Four Variables $\left(3(xyz+yzt+ztx+txy)^2\ge 8(xy+xz+xt+yz+yz+zt)xyzt\right)$
- Dan Sitaru's Cyclic Inequality in Four Variables $\left(\displaystyle \sum_{cycl}\frac{a^7}{bcd+a^3}\ge 2abcd\right)$
- A Not Quite Cyclic Inequality from Tibet $\left((x+y)^2(z^2+xz+x^2+xy+y^2+yz)^2\ge 8(xy+yz+zx)^2(x^2+y^2)\right)$
- An inequality in 2+2 variables from SSMA magazine $\left(\displaystyle k^2\tan\alpha+l^2\tan\beta\ge\frac{2kl}{\sin (\alpha+\beta)}-(k^2+l^2)\cot(\alpha+\beta)\right)$
- Kunihiko Chikaya's Inequality with Parameter $\bigg(p \ge 2.$ Prove $\displaystyle \sum_{cycl}\frac{a}{\sqrt{ap+b}} \le \sqrt{\frac{3(a+b+c)}{p+1}}\bigg)$
- Dorin Marghidanu's Permuted Inequality $\left(\displaystyle \sum_{k=1}^n\left(a_k+\frac{1}{a_{\sigma(k)}}\right)^p\ge \frac{(s^2+n^2)^p}{n^{p-1}s^p}\right)$
- An Inequality Involving Arithmetic And Geometric Means $\left(\displaystyle\sum_{cycl}\frac{1}{a^4+b^4+c^4+abcd}\le \frac{1}{abcd}\right)$
- Dorin Marghidanu's Sums and Products $\left(\displaystyle \sum_{k=1}^n\frac{a_k}{P_kS_k}\ge\frac{n^n}{\displaystyle (n-1)S^{n-1}}\right)$
- Simple Nameless Inequality $\left(\displaystyle \sum_{k=1}^n\frac{S}{S_k}\ge\frac{n^2}{n-1}\right)$
- Volume Inequality in Tetrahedron $\left(OA\cdot OB\cdot OC\ge 27xyz\right)$
- Inequality in Convex Quadrilateral $\left(\displaystyle\frac{\displaystyle \sum_{cycl}\sqrt{b+c+d-a}}{a+b+c+d}\ge\sqrt{\frac{2(a+b+c+d)}{a^2+b^2+c^2+d^2}}\right)$
- Dan Sitaru's Inequality with a Double Integral $\left(\displaystyle\begin{align}&\small{\int_0^1\int_0^1\sqrt{\left(m^2\sqrt{mnf(x)f(y)}+f^2(x)\right)\left(n^2\sqrt{mnf(x)f(y)}+f^2(y)\right)}dxdy}\\ &\qquad\qquad\qquad\qquad\small{\ge (m+n)\int_0^1f(x)dx}.\end{align}\right)$
- Cute Exercise by Dorin Marghidanu $\left(\displaystyle \sum_{k=1}^n\frac{2k-1}{\sqrt[2k-1]{a_k}}\ge\frac{n^2}{\sqrt[n^2]{a_1a_2\ldots a_n}}\right)$
- A Little of Algebra for an Inequality, A Little of Calculus for a Generalization $\left(\displaystyle \frac{a^{n+1}-b^{n+1}}{a-b}\cdot\frac{b^{n+1}-c^{n+1}}{b-c}\cdot\frac{c^{n+1}-a^{n+1}}{c-a}\gt (n+1)^3(abc)^n\right)$
- An Inequality with Central Binomials $\left(\displaystyle \sqrt{2}\le\sqrt[n(n+1)]{{2\choose 1}{4\choose 2}\cdots{2k\choose k}\cdots{2n\choose n}}\lt 2\right)$
- A Simple Inequality with Many Variables $\left(\displaystyle \sum_{k=1}^n\sqrt{\frac{a_k+a_{k+1}}{a_{k+2}}}\ge n\sqrt{2}\right)$
- Cyclic Inequality in Four Variables $\left(\displaystyle \frac{3}{4}\sum_{cycl}\frac{a^3}{bcd}\ge 1+\frac{\displaystyle 3\sum_{cycl}a^2}{\displaystyle \sum_{all}ab}\right)$
- Cyclic Inequality in Four Variables By D. Sitaru $\left(\displaystyle \sum_{cycl}\frac{a^7}{a^3+bcd}\ge 2abcd\right)$
- Lorian Saceanu's Inequality with Many Variables $\left(\displaystyle \frac{1}{2}\left(\sqrt{ab}+\frac{1}{\sqrt{ab}}\right)\left(\sum_{i=1}^na_kb_k\right)\ge\sqrt{\left(\sum_{i=1}^na_k^2\right)\left(\sum_{i=1}^nb_k^2\right)}\right)$
- A True Algebraic-Geometric Inequality $\left(\displaystyle \small{\sqrt{\sum_{k=1}^n(2a_k-b_k)^2}+\sqrt{\sum_{k=1}^n(2b_k-a_k)^2}\ge\sqrt{\sum_{k=1}^na_k^2}+\sqrt{\sum_{k=1}^nb_k^2}}\right).$
- Leo Giugiuc's Cyclic Inequality in Square Roots $\left(\displaystyle \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge 2\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}\right)$
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