$\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge $ $\frac{3}{2} - \frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{12}}$

Let $x>0, y>0, z>0$ satisfy $x+y+z=3$. Prove that,

$\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2} - \frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{12}}$

Solution

Using Titu Andreescu's lemma,

$\frac{{a_1^2}}{{{b_1}}} + \frac{{a_2^2}}{{{b_2}}} + ............. + \frac{{a_n^2}}{{{b_n}}} \ge \frac{{{{({a_1} + {a_2} + ........ + {a_n})}^2}}}{{{b_1} + {b_2} + ....... + {b_n}}}$

for ${b_i} > 0$

We have, $\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{{{{(x + y + z)}^2}}}{{(1 + x) + (1 + y) + (1 + z)}}$

$ \Rightarrow \frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{{{{(x + y + z)}^2}}}{{3 + (x + y + z)}}$

$ \Rightarrow \frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2} \because x + y + z = 3$

The expression $\frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{12}}$ is non-negative. So, if

$\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2}$ then $\frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2} - ( + ve)$

$\therefore \frac{{{x^2}}}{{1 + x}} + \frac{{{y^2}}}{{1 + y}} + \frac{{{z^2}}}{{1 + z}} \ge \frac{3}{2} - \frac{{{{(x - y)}^2} + {{(y - z)}^2} + {{(z - x)}^2}}}{{12}}$