Hint

Is this from Pham Kim hungs book?

trungnghia215 wrote:

Hint

Writing similar inequalities & substituting we get $-\frac{x+y+z}{2}+3\geq 3$ so? then we get $-1,5\geq 0$ o.O

Ok, so split LHS terms like 2/1+x as 1/1+x + 1/1+x And then apply titu's lemma

Imgoodinmaths_dude wrote: trungnghia215 wrote:

Hint

Writing similar inequalities & substituting we get $-\frac{x+y+z}{2}+3\geq 3$ so? then we get $-1,5\geq 0$ o.O Serious?... $f(a) + f(b) + f(c) \ge -\frac{a + b + c}{2} + \frac{9}{2} \ge 3$.

Alright.... i wrote $\frac{3}{2}*3=3$ :/

Titu's to the rescue

trungnghia215 wrote:

Hint

From Cauchy Schwarz we have $LHS\ge 2.(\frac{9}{3+x+y+z})\ge 2.(\frac{9}{3+3})=3$

My Solution: By Titu form of C-S, $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z} \geq \frac{9}{3+x+y+z} \geq \frac{9}{6} = \frac{3}{2}$ which implies the result.

Titu's lemma kills this question :/

$x+y+z \leq 3$ $\frac{(x+1)+(y+1)+(z+1)}{3} \leq 2$ By AM-HM $\frac{3}{\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}} \leq \frac{(x+1)+(y+1)+(z+1)}{3} \leq 2$ So, $\frac{2}{1+x}+\frac{2}{1+y}+\frac{2}{1+z} \geq 3$

trivial by Cauchy-Engel or Jensen's Inequality

AM-HM on $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z}$ proves the stated inequality