# An Inequality from Gazeta Matematica, March 2016 III

### Problem

Several inequalities with solution by Dan Sitaru and Leo Giugiuc have been just published in Gazeta Matematica (March 2016). Here is one with two of its applications and a proof (Proof 1)from the article. Along the way several additional proofs have been added. Proof 2 is by Imad Zak; Proof 3 is by Emil Stoyanov; Proof 4 is by Grégoire Nicollier; Proof 5 is by Kunihiko Chikaya; Proofs 6 and 7 are by Robert Kosova; Proof 8 by Abdur Rahman.

(Marco Antônio Manetta has observed that the requirement that the three numbers are positive is redundant.)

### Proof 1

Define $A=\left(\begin{array}\;1&a&b\\a&b&1\end{array}\right).$ By the Binet-Cauchy theorem, $\det(AA^T)\ge 0.\;$ But

$\det(AA^T)=(a^2+b^2+1)^2-(a+ab+b)^2,$

proving the inequality at hand.

### Proof 2

Let $S=a+b\;$ and $P=ab,\;$ by the AM-GM inequality, we have $\displaystyle P\le \frac{S^2}{4}\;$ and the required inequality is equivalent to $S^2-S+1\ge 3P,\;$ so suffice it to prove that $\displaystyle S^2-S+1\ge\frac{3S^2}{4}\;$ which is equivalent to $\displaystyle\frac{S^2}{4}-S+1\ge 0,\;$ or $\displaystyle (\frac{S}{2}-1)^2\ge 0\;$ which is clearly true. The equality holds when $S=2\;$ and $P=1,\;$ i.e., when $a=b=1.$

### Proof 3

The required inequality is equivalent to $a^2-(b+1)a+(b^2-b+1)\ge 0.\;$ Consider the quadratic function $f(x)=x^2-(b+1)x+(b^2-b+1)\ge 0.\;$ Its discriminant $D=(b+1)^2-4b^2+4b-4=-3(b-1)^2\;$ is never positive, implying that function $f\;$ is never negative.

### Proof 4

The inequality reduces to to $(a-1)^2 + (b-1)^2 \ge (a-1)(b-1)\;$ which could be strengthened to $(a-1)^2 + (b-1)^2 \ge 2(a-1)(b-1).$

### Proof 5

The required inequality is equivalent to

$(a+b-2)^2+3(a-b)^2\ge 0.$

### Proof 6

The inequality is equivalent to $2a^2+2b^2+2\ge 2a+2b+2ab.$

But $2a\le a^2+1,\;$ $2b\le b^2+1.\;$ Thus suffice it to show that

$2a^2+2b^2+2\ge (a^2+1)+(b^2+1) +2ab,$

or, $a^2+b^2\ge 2ab,\;$ which is $(a-b)^2\ge 0,\;$ and is true.

### Proof 7

\begin{align} a^2+b^2 &\ge 2ab\\ a^2+1 &\ge 2a\\ b^2+1 &\ge 2b. \end{align}

Adding up gives $2a^2+2b^2+2\ge 2a+2ab+2b,\;$ and we only have to divide by $2.$

### Proof 8

By the Cauchy-Schwarz inequality,

\begin{align} (a+ab+b)^2 &= (a\cdot 1+ b\cdot a+1\cdot b)\\ &\le (a^2+b^2+1)(1+a^2+b^2), \end{align}

implying $a+ab+b\le a^2+b^2+1.$

### Proof 9

By the AM-QM inequality,

$\displaystyle a^2+b^2+1\ge \frac{1}{2}(a+b)^2+1.$

Suffice it to prove that

$(a+b)^2+2\ge 2a+2ab+2b.$

But this is equivalent to $(a-1)^2+(b-1)^2\ge 0,\;$ which is obvious.

### Application 1

$\displaystyle\prod_{1\le i\le j\le n}(i^2+j^2+1)\ge n!\prod_{1\le i\le j\le n}(2+\sqrt{ij}).$

Observe that $a^2+b^2+1\ge a+b+ab\ge ab+2\sqrt{ab}=\sqrt{ab}(2+\sqrt{ab}).\;$ Using this,

\displaystyle\begin{align} \prod_{1\le i\le j\le n}(i^2+j^2+1) &\ge \prod_{1\le i\le j\le n}\sqrt{ij}(2+\sqrt{ij})\\ &=\prod_{1\le i\le j\le n}\sqrt{ij}\prod_{1\le i\le j\le n}(2+\sqrt{ij})\\ &=n!\prod_{1\le i\le j\le n}(2+\sqrt{ij}). \end{align}

Obviously, the inequality can be strengthened.

### Application 2

Show that $\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin x+\sin x\cos x+\cos x}dx\ge\frac{\pi}{4}.$

Set $a=\sin x\;$ and $b=\cos x.\;$ Then $2\ge 1+\sin^2x+\cos^2x\ge \sin x+\sin x\cos x+\cos x,\;$ implying

$\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin x+\sin x\cos x+\cos x}dx \ge\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\ge\frac{\pi}{4}.$

Note that, according to wolframalpha, $\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin x+\sin x\cos x+\cos x}dx\approx 1.02245.$

### Linear Algebra Tools for Proving Inequalities

$\;\left(\displaystyle\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2\ge 2\right)$
• Linear Algebra Tools for Proving Inequalities: Cauchy-Binet Formula $\;\left(\displaystyle\left(\sum_{i=1}^{n}\frac{x_i^2}{a_i}\right)\cdot\left(\sum_{1\le i\lt j\le n}a_ia_j(x_iy_j-x_jy_i)^2\right)\ge \sum_{i=1}^{n}a_iy_i^2\right)$
• An Inequality from Gazeta Matematica, March 2016 (If $a^2+b^2+c^2=3\,$ then $(a+c)(1+b)\le 4)$
• An Inequality from Gazeta Matematica, March 2016 II (If $x^2+y^2+z^2+t^2=1\,$ then $\;(x+z)(y+t)\le 4)$
• An Inequality from Gazeta Matematica, March 2016 IV (If $a^2+b^2+c^2=1\,$ then $a+ac+b\le 2)$
• Problem 3980 from Crux Mathematicorum $\;\left(\displaystyle\sum_{cycl}\frac{a+b}{a-b}\prod_{cycl}\frac{a+b}{a-b}\lt\frac{1}{3}\right)$
• NonSquare Matrix as a Tool for Proving an Inequality $\;\left(2(a + b + c)((a + 2b + 3c) \ge (\sqrt{b(a+b)} + 2\sqrt{c(b+c)} + \sqrt{a(c+a)})^2\right)$
• An Inequality in Parallelogram of Unit Area $\;\left(a^2+b^2+c^2+d^2+ac+bd\ge\sqrt{3}\right)$
• An Inequality from a Vietnamese Problem Book $\;\left(\displaystyle \frac{a^3+2}{b+2c}+\frac{b^3+2}{c+2a}+\frac{c^3+2}{a+2b}\ge 3\right)$
• Hadamard's Determinant Inequalities and Applications I $\left((2-a-b-c+abc)^2\le (a^2+2)(b^2+2)(c^2+2)\right)$
• Hadamard's Determinant Inequalities and Applications II $\left((n + a - 1)(a - 1)^{n-1} \le a^n\right)$
• Copyright © 1996-2018 Alexander Bogomolny

70192676