# work 015 – Pham Kim Hung

Prove that for $a,b,c$ side-lengths of a triangle then prove that the following inequality holds ,

$\sum (2+\frac{a}{b})^2 \ge \frac{9(a+b+c)^2}{ab+bc+ca}$

Note : The inequality as proposed originally holds for all positive reals….but
I amn’t good enough to solve it without the extra condition I have imposed.

Proof

Lemma 1

For all positive reals ,

$\sum \frac{a}{b} \ge 3\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$

Lemma 2

For all sidelengths of a triangle ,

$\sum \frac{a+b-2c}{b+c} \ge 0$

Now,

By expanding the LHS and using Lemma 1 we only have to prove that –

$2\sum \frac{a}{b} \ge \sum \frac{a}{c} + 3$

Since they $a,b,c$ are side-lengths of a triangle , using Ravi’s Substitution ,

the inequality re-arranges to –

$\sum \frac{x+y-2z}{y+z} \ge 0$

where , $x+y=a , y+z=b , z+x=c$

which is true by Lemma 2.

This completes the proof.