Dan Sitaru's Cyclic Inequality In Three Variables with Constraints IV

Problem

Dan Sitaru's Cyclic  Inequality In  Three Variables with Constraints IV, problem

Solution 1

First of all, we simplify the problem by replacing the variables: $a=x,\,$ $b=2y,\,$ $c=3z,\,$ which reduces the problem to

Prove that if $a,b,c\gt 0;\,abc(a+b+c)=1\,$ then:

$\displaystyle \frac{(a^2b^2+1)(b^2c^2+1)(c^2a^2+1)}{a^2b^2c^2}\geq \frac{64}{(a+b+c)^2}.$

We rewrite the inequality as

$\displaystyle \left(a^2+\frac{1}{b^2}\right)\left(b^2+\frac{1}{c^2}\right)\left(c^2+\frac{1}{a^2}\right)\geq \frac{64}{(a+b+c)^2}.$

Note that

$\displaystyle \begin{align} a^2+\frac{1}{b^2} &= a^2+\frac{abc(a+b+c)}{b^2}=a^2+\frac{ac(a+b+c)}{b}\\ &=\frac{a^2b+a^2c+ac(b+c)}{b}=\frac{a^2(b+c)+ac(b+c)}{b}\\ &=\frac{a(b+c)(c+a)}{b}. \end{align}$

Similarly $\displaystyle b^2+\frac{1}{c^2}=\frac{b(c+a)(a+b)}{c}\,$ and $\displaystyle c^2+\frac{1}{a^2}=\frac{c(a+b)(b+c)}{a}.$

By multiplying the three relationships,

$\displaystyle \begin{align} &\Bigr(a^2+\frac{1}{b^2}\Bigr)\Bigr(b^2+\frac{1}{c^2}\Bigr)\Bigr(c^2+\frac{1}{a^2}\Bigr)\\ &\qquad\qquad\geq \frac{a(b+c)(a+c)\cdot b(a+c)(b+a)\cdot c(a+b)(c+b)}{abc}\\ &\qquad\qquad=(a+b)^2(b+c)^2(c+a)^2\\ &\qquad\qquad\overbrace{\geq}^{AM-GM}(2\sqrt{ab})^2\cdot (2\sqrt{bc})^2 \cdot (2\sqrt{ac})^2\\ &\qquad\qquad=64a^2b^2c^2=\frac{64}{(a+b+c)^2} \end{align}$

The equality holds if $\displaystyle a=b=c=\frac{1}{\sqrt[4]{3}},\,$ which follows from

$\displaystyle a\cdot a\cdot a =\frac{1}{a+a+a}\Leftrightarrow a^3=\frac{1}{3a}\Rightarrow a^4=\frac{1}{3}\Rightarrow a=\frac{1}{\sqrt[4]{3}}.$

The equality in the original relationship holds for

$\displaystyle x=\frac{1}{\sqrt[4]{3}}; y=\frac{1}{2\cdot\sqrt[4]{3}}; z=\frac{1}{3\cdot\sqrt[4]{3}}$

Solution 2

Using the constraint, the inequality can be written as

$\displaystyle (4x^2y^2+1)(36y^2z^2+1)(9x^2z^2+1)\geq 64(6xyz)^4$

Let,

$\displaystyle x=\sqrt{\frac{ab}{c}},~2y=\sqrt{\frac{bc}{a}},~3z=\sqrt{\frac{ca}{b}}.$

Thus, the inequality and the constraint, respectively, become

$\displaystyle (a^2+1)(b^2+1)(c^2+1)\geq64(abc)^2,~ab+bc+ca=1.$

Let $p=3(abc)^{2/3}$. AM-GM gives

$\displaystyle 1=ab+bc+ca\geq3(abc)^{2/3}=p.$

The inequality can be simplified to

$\displaystyle 1+(a^2+b^2+c^2)+(a^2b^2+b^2c^2+c^2a^2)-63(abc)^2\geq 0.$

$\displaystyle \begin{align} LHS &\geq 1+3(abc)^{2/3}+3(abc)^{4/3}-63(abc)^2~\text{(AM-GM)} \\ &= 1+p+\frac{p^2}{3}-\frac{7}{3}p^3\\ &= \frac{(1-p)}{3}[7(1-p)^2-20(1-p)+16] \\ &= \frac{(1-p)}{3}\left\{\left[(1-p)\sqrt{7}-\frac{10}{\sqrt{7}}\right]^2+\frac{12}{7}\right\}\geq 0, \end{align}$

from the constraint.

Acknowledgment

Dan Sitaru has kindly posted the above problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page and later emailed me his solution. Solution 2 is by Amit Itagi.

A refinement of the inequality suggested by N. N. Taleb is dealt with on a separate page.

 

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