An Inequality from Gazeta Matematica, March 2016


An inequality with two solutions by Dan Sitaru and Leo Giugiuc has been just published in Gazeta Matematica (March 2016).

an inequality from Gazeta Matematica, March 2016

Proof 1

Define matrix $\displaystyle A=\left(\begin{array}\;1&a\\a&b\\b&c\\c&1\end{array}\right).\;$ We have

$\displaystyle\begin{align} A^t\cdot A &= \left(\begin{array}\;1&a&b&c\\a&b&c&1\end{array}\right)\cdot \left(\begin{array}\;1&a\\a&b\\b&c\\c&1\end{array}\right)\\ &=\left(\begin{array}\;a^2+b^2+c^2+1 & a+ab+bc+c\\a+ab+bc+c & a^2+b^2+c^2+1\end{array}\right)\\ &=\left(\begin{array}{c,c}\;4 & (a+c)(1+b)\\(a+c)(1+b) & 4\end{array}\right). \end{align}$

By Cauchy-Binet theorem, $\text{det}(A^t\cdot A)\ge 0.\;$ Therefore, $[(a+c)(1+b)]^2\le 16,\;$ or, $(a+c)(1+b)\le 4.$

Proof 2

We use spherical coordinates. Let $b=\sqrt{3}\cos t,\;$ $a=\sqrt{3}\sin t \cos u,\;$ and $c=\sqrt{3}\sin t \sin u,\;$ where $\displaystyle 0\lt t\lt\frac{\pi}{2}.\;$ We need to prove that $\sqrt{3}(\cos u+\sin u)(1+\sqrt{3}\cos t)\sin t\le 4.$

Observe that $1\lt \sin u+\cos u\le\sqrt{2}.\;$ Thus, suffice it to prove that $\sqrt{6}(1+\sqrt{3}\cos t)\sin t\le 4.$

Consider the function $\displaystyle f;\,\left(0,\frac{\pi}{2}\right)\rightarrow\mathbb{R},\;$ defined by $f(t)=(1+\sqrt{3}\cos t)\sin t.\;$ We have $f'(t)=2\sqrt{3}\cos^2t+\cos t-\sqrt{3}\sin t\;$ which implies $\displaystyle \text{max}f=f\left(\arccos\left(\frac{1}{\sqrt{3}}\right)\right)=2\sqrt{\frac{2}{3}}.\;$ Therefore, $\displaystyle \sqrt{6}(1+\sqrt{3}\cos t)\sin t\le\sqrt{6}\cdot 2\sqrt{\frac{2}{3}}=4.$

Proof 3

$(a-b)^2+(b-c)^2+(a-1)^2+(c-1)^2\ge 0\;$ which simplifies to $(a^2+b^2+c^2)+1-ab-bc-a-c\ge 0.\;$ This is exactly $(a+c)(1+b) \le 4.$

Proof 4

By the AM-QM inequalty, $\displaystyle\frac{a+b+c}{3}\le\sqrt{\frac{a^2+b^2+c^2}{3}}=1.\;$ Further, by the AM-GM inequality,

$\displaystyle\begin{align} (a+c)(b+1) &\le \left(\frac{a+c+b+1}{2}\right)^2\\ &\le \left(\frac{3+1}{2}\right)^2\\ &=4. \end{align}$

Proof 5

From $a^2+b^2+c^2=3,\;$ $3\sum a^2\ge\left(\sum a\right)^2,\;$ implying $\left(\sum a\right)^2\le 9,\;$ or, $a+b+c\le 3.\;$ $a+b+c+1\le 4.\;$ Therefore,

$4\ge (a+c)+(b+1)\ge 2\sqrt{(a+c)(b+1)},\;$ i.e., $2\ge \sqrt{(a+c)(b+1)},\;$ or $4\ge (a+c)(b+1).$

Equality is attained when $a+c=b+1,\;$ which, with $a^2+b^2+c^2=3\;$ implies $a=b=c=1.$

Proof 6

From $(x-y)^2\ge 0\;$ we have $2xy\le x^2+y^2.\;$ We use this with the couples $(a,b),\;$ $(b,c),\;$ $(1,a),\;$ $(1,c):\;$

$2ab\le a^2+b^2\\ 2bc\le b^2+c^2\\ 2a\le 1+a^2\\ 2c\le 1+c^2$

adding which gives $2(a+c+ab+bc)\le 2+2(a^2+b^2+c^2)=8,\;$ and this is exactly $(a+c)(b+1)\le 4.$

Proof 7

A simple inequality from Gazeta Matematica, proof by Kunihiko Chikaya

To continue:

$\displaystyle\begin{align} (a+c)(1+b) &\le \sqrt{2}\sqrt{3-b^2}(1+b)\\ &\le\sqrt{2}\sqrt{4-2b})1+b)\\ &=\sqrt{2}\sqrt{(4-2b)(1+b)^2}\\ &=\sqrt{2}\sqrt{(1+b)(1+b)(4-2b)}\\ &\le\sqrt{2}\sqrt{\left(\frac{1+b+1+b+4-2b}{3}\right)^3}\\ &=4. \end{align}$

The equality is achieved for $1+b=1+b=4-2b,\;$ making $b=1\;$ which, by the way, satisfies $3-b^2\ge 0.\;$ $\displaystyle a=c=\sqrt{\frac{3-b^2}{2}},\;$ i.e., $a=b=c=1.$

Proof 8

A simple inequality from Gazeta Matematica, second proof by Kunihiko Chikaya

To continue:

$\displaystyle\begin{align} (a+c)(1+b) &\le \sqrt{2}\sqrt{3-b^2}(1+b)\\ &\le\sqrt{2}\frac{3-b}{\sqrt{2}}(1+b)\\ &\le\left(\frac{3-b+1+b}{2}\right)^2\\ &=4. \end{align}$

Equality is achieved for $3-b=1+b\;$ and $\displaystyle a=c=\sqrt{\frac{3-b^2}{2}},\;$ i.e., $a=b=c=1.$

Proof 9

Observe that $(a+c)(1+b)=a\cdot 1+a\cdot b+c\cdot 1+b\cdot c\;$ such that by the Cauchy-Schwarz inequality,

$(a\cdot 1+a\cdot b+c\cdot 1+b\cdot c)^2 \le (a^2+b^2+c^2+b^2)(1^2+a^2+1^2+c^2)$

which leads to a chain of inequalities

$\displaystyle\begin{align} (a+ab+c+bc)^2 &\le (3+b^2)(2+a^2+c^2)\\ \left[(a+c)(1+b)\right]^2 &\le (3+b^2)(2+3-b^2)\\ &= (3+b^2)(5-b^2)\\ &\le \left[\frac{3+b^2+5-b^2}{2}\right]^2\\ &= \left[\frac{8}{2}\right]^2, \end{align}$

and, therefore, $(a+c)(1+b)\le 4.$

Proof 10

From $(a+c)^2\le 2(a^2+c^2)\;$ and $(1+b)^2\le 2(1+b^2)\;$ we obtain a sequence of inequalities:

$\displaystyle\begin{align} (a+c)^2(1+b)^2 &\le 4(a^2+c^2)(1+b^2)\\ &\le 4\left(\frac{a^2+c^2+1+b^2}{2}\right)^2\\ &= 4\left(\frac{4}{2}\right)^2\\ &= 4\cdot 4. \end{align}$

and, therefore, $(a+c)(1+b)\le 4.$

Proof 11

We prove $((a+b)(1+c))^2\le 16\;$ when the point $(a,b,c)\;$ lies on the sphere of radius $\sqrt{3}\;$ centered at the origin. At height $c\;$ the sphere is a circle of radius $r=\sqrt{3-c^2}\;$ and the maximum of $a+b\;$ is $r\sqrt{2}\;$ (consider the line with slope $-1\;$ tangent to this circle in the first quadrant of the plane). We want thus the maximum of $2(1+c)^2 (3-c^2)\;\;$ for $0 \le c \le \sqrt{3}.\;$ The value $16\;$ is attained for $c=1.\;$ But $16 - 2(1+c)^2 (3-c^2) = 2(c-1)^2 (5 + 4 c + c^2)\ge 0,\qquad$ for all real $c.$

Proof 12

Use Lagrange multipliers to prove that



Let $J=(a+c)(1+b)+\lambda (a^2+b^2+c^2-3).$

Taking $\displaystyle\frac{\partial J}{\partial a}=\frac{\partial J}{\partial b}=\frac{\partial J}{\partial c}=\frac{\partial J}{\partial\lambda}=0\;$ yields










$a=c\;$ (from (2) and (4))


$2a^2+2ab\lambda=0\;$ (from (3) and(6))


$b+b^2+2ab\lambda=0\;$ (from (2))


$b+b^2=2a^2\;$ (from (7) and (8))


$2a^2+b^2=3$ (from (5) and (6))


$2b^2+b-3=0,\; b=1,-3/2$ (from (9) and (10))

Hence, $b=1\;$ and from (10), $a=\pm 1,\;$ implying $a=c=1\;$ and $b=1,\;$ and (1) follows.


It is clear that the equality is attained for $a=b=c=1\;$ - a symmetric condition whereas the inequality itself is asymmetric. In analogy with the above derivation, we can show $(b+a)(1+c)\le 4\;$ and $(b+c)(1+a)\le 4.\;$ The sum of the three gives $(a+b+c)+(ab+bc+ca)\le 6\;$ which is just more symmetric.


Proofs 1 and 2 are by Leo Giugiuc and Dan Sitaru; Proof 3 is by Nevena Sybeva; Proof 4 is by Augustini Moraru; Proof 5 is by Imad Zak and independently by Rahim Shahbazov; Proof 6 is by Robert Kosova; Proofs 7 and 8 are by Kunihiko Chikaya; Proofs 9 and 10 are by Sk Rejuan; Proof 11 is by Grégoire Nicollier; Proof 12 is by Michalos Nikolau. My sincerest gratitud goes to Leo Giugiuc for furnishing this collection.


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