# An Inequality in Triangle III

### Problem

Prove that in $\Delta ABC,\;$ with angles $A,B,C\;$ side lengths $a,b,c\;$ the following inequality holds:

### Solution 1

WLOG, assume $a\ge b\ge c.\;$ Then $ab+ac\ge bc+ca\ge cb+ca;\;$ also, $\displaystyle\frac{1}{bc}\ge\frac{1}{ca}\ge\frac{1}{ab}.\;$ From these, $\displaystyle\frac{ab+ac}{bc}\ge\frac{ab+bc}{ac}\ge\frac{bc+ac}{ab}.\;$ On the other hand, $\displaystyle\frac{1}{\cos^2 x}=1+\tan^2x\;$ and the tangent function in strictly increasing and positive on $\displaystyle(0,\frac{\pi}{2},\;$ hence $\displaystyle 1+\tan^2\frac{A}{2}\ge 1+\tan^2\frac{B}{2}\ge 1+\tan^2\frac{C}{2}.\;$ We can apply now Chebyshev's inequality to get

$\displaystyle\sum_{cyc}\frac{ab+ac}{\displaystyle bc\cdot\cos^2\frac{A}{2}}\ge\frac{1}{3}\left(\frac{ab+ac}{bc}+\frac{ab+bc}{ac}+\frac{bc+ac}{ab}\right)\left(3+\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\right).$

Suffice it to prove that

$\displaystyle\frac{1}{3}\left(\frac{ab+ac}{bc}+\frac{ab+bc}{ac}+\frac{bc+ac}{ab}\right)\left(3+\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2}\right)\ge 24.$

But obviously $\displaystyle\frac{ab+ac}{bc}+\frac{ab+bc}{ac}+\frac{bc+ac}{ab}\ge 6.\;$ On the other hand,

\displaystyle\begin{align} 3+\tan^2\frac{A}{2}+\tan^2\frac{B}{2}+\tan^2\frac{C}{2} &\ge 3+\frac{1}{3}\left(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\right)^2\\ &\ge 3 +1=4. \end{align}

### Solution 2

From the half-angle formula an the Law of Cosine,

$\displaystyle\cos^2\frac{A}{2}=\frac{1+\cos A}{2}=\frac{(b+c)^2-a^2}{4bc},$

and similarly for the other two angles. Thus the inequality at hand is equivalent to

$\displaystyle\frac{a(b+c)}{(b+c)^2-a^2}+\frac{b(c+a)}{(c+a)^2-b^2}+\frac{c(a+b)}{(a+b)^2-c^2}\ge 2,$

or,

$\displaystyle\frac{a(b+c)}{b+c-a}+\frac{b(c+a)}{c+a-b}+\frac{c(a+b)}{a+b-c}\ge 2(a+b+c),$

or,else

$\displaystyle\frac{a^2}{b+c-a}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}\ge a+b+c,$

Now, by the Cauchy-Schwarz inequality, for any $x,y,z\gt 0,$

$\displaystyle\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge \frac{(a+b+c)^2}{x+y+z},$

which, with $x=b+c-a,\;$ $y=c+a-b,\;$ $z=a+b-c\;$ gives

$\displaystyle\frac{a^2}{b+c-a}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}\ge \frac{(a+b+c)^2}{a+b+c}=a+b+c.$

### Solution 3

We know that $\displaystyle\cos^2\frac{A}{2}=\frac{p(p-a)}{bc},\;$ where $p=(a+b+c)/2\;$ is the semiperimeter of $\Delta ABC.\;$ Thus, the required inequality can be rewritten as

$\displaystyle\frac{a(b+c)}{p(p-a)}+\frac{b(c+a)}{p(p-b)}+\frac{c(b+a)}{p(p-c)}\ge 8.$

Now observe that $\displaystyle\frac{a(b+c)}{p(p-a)}=\frac{2a}{p}+\frac{a^2}{p(p-a)},\;$ and similar for the other two fractions. Since, $\displaystyle\sum_{cyc}\frac{2a}{p}=4,\;$ the required inequality reduces to

$\displaystyle\frac{a^2}{p(p-a)}+\frac{b^2}{p(p-b)}+\frac{c^2}{p(p-c)}\ge 4.$

Consider the function $\displaystyle f(x)=\frac{x^2}{p-x}.\;$ $\displaystyle f'(x)=\frac{2xp}{(p-x)^2}\gt 0\;$ and $\displaystyle f''(x)=\frac{2p(p+x)}{(p-x)^3}\lt 0\;$ for $x\in (0,p).\;$ Thus the function is convex on $(0,p).\;$ Keeping $p\;$ fixed, we may apply Jensen's inequality:

\displaystyle\begin{align} \frac{a^2}{p(p-a)}+\frac{b^2}{p(p-b)}+\frac{c^2}{p(p-c)}&\ge 3\frac{\displaystyle\left(\frac{a+b+c}{3}\right)^2}{p\left(p-\displaystyle\frac{a+b+c}{3}\right)}\\ &=3\frac{\displaystyle\left(\frac{2p}{3}\right)^2}{p\left(p-\displaystyle\frac{2p}{3}\right)}\\ &=3\cdot\frac{4}{9}\cdot\frac{3}{1}\\ &= 4. \end{align}

### Solution 4

\displaystyle \begin{align} \sum_{cycl}\frac{a(b+c)}{\displaystyle bc\cdot\cos\frac{A}{2}}&=\sum_{cycl}\frac{a(b+c)}{\displaystyle bc\cdot\frac{p(p-a)}{bc}}\\ &=\sum_{cycl}\frac{a(b+c)}{p(p-a)}=\sum_{cycl}\frac{a(2p-a)(p-b)(p-c)}{p(p-a)(p-b)(p-c)}\\ &=\frac{4R}{r}\ge 8, \end{align}

due to Euler's inequality $R\ge 2r.$

### Acknowledgment

The problem from the Math Phenomenon has been posted at the CutTheKnotMath facebook page by Dan Sitaru, along with a solution (Solution 1) by Leo Giugiuc and Dan Sitaru. Solution 2 is by Kunihiko Chikaya; Solution 4 is by Marin Chirciu.