Prove that $\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}$

Question

If $a+b+c=3$ ,$ a,b,c>0$ and $\lambda \geq 1$, prove that : $$\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}.$$

my attempt:

using CBC twice and AM_AG inequality

$\frac{3}{3}\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)^2\leq 3 \Big(\frac{a}{a^2+\lambda}\Big)^2+\Big(\frac{b}{b^2+\lambda}\Big)^2+\Big(\frac{c}{c^2+\lambda}\Big)^2\leq 3\Bigg( \bigg(\frac{a}{\frac{(a+\sqrt{\lambda})^2}{2}}\bigg)^2+ \bigg(\frac{b}{\frac{(b+\sqrt{\lambda})^2}{2}}\bigg)^2+\bigg(\frac{c}{\frac{(c+\sqrt{\lambda})^2}{2}}\bigg)^2\Bigg) =12\bigg(\frac{a^2}{(a+\sqrt{\lambda })^4}+\frac{b^2}{(b+\sqrt{\lambda})^4}+\frac{c^2}{(c+\sqrt{\lambda })^4}\bigg)\leq12 \bigg(\frac{a^2}{\sqrt{\lambda}^4}+\frac{b^2}{\sqrt{\lambda}^4}+\frac{c^2}{\sqrt{\lambda}^4}\bigg)=\frac{48(a^2+b^2+c^2)}{4\lambda ^2 }\leq \frac{48\cdot9}{4\lambda^2} \leq \frac{9}{(\lambda +1)^2}$

beacuse $f(\lambda )=\frac{9}{(\lambda+1)^2}-\frac{48\cdot9}{4\lambda^2} \geq 0 $ for $\lambda \geq 1$

so finally $\frac{3}{3}\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)^2\leq\frac{9}{(\lambda +1)^2}$$\Leftrightarrow$$\Big(\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda}\Big)\leq\frac{3}{(\lambda +1)}$

I have just one question:

-does my attempt is true?

Answer 1

Tangent Line (TL) method:

It suffices to prove that $$\frac{a}{a^2 + \lambda} \le \frac{1}{1 + \lambda} + \frac{\lambda - 1}{(1 + \lambda)^2}(a - 1). \tag{1}$$ (The desired result follows by summing cyclically.)

(1) is true since $$\frac{a}{a^2 + \lambda} - \frac{1}{1 + \lambda} - \frac{\lambda - 1}{(1 + \lambda)^2}(a - 1) = - \frac{(a - 1)^2(a\lambda - a + 2\lambda)}{(a^2 + \lambda)(1 + \lambda)^2} \le 0.$$

We are done.

Answer 2

A comment on the tangent line method. Why does it work? The function $f$ is not concave everywhere. What's going on?

The function $f_{\lambda}=f\colon x\mapsto \frac{x}{x^2 + \lambda}$ has derivatives $$f'(x) = \frac{\lambda - x}{(\lambda + x^2)^2} \\ f^{''}(x) = \frac{2x(x^2 -3 \lambda) }{(x^2 + \lambda)^3}$$

Therefore, $f$ is strictly increasing on the interval $[0, \sqrt{\lambda}]$ and strictly concave on the interval $[0, \sqrt{3 \lambda}]$. We conclude that the function $$f_1(x) = \begin{cases} f(x) &\textrm{ for }& x \in [0, \sqrt{\lambda}] \\ f(\sqrt{\lambda}) &\textrm{ for }& x \in [\sqrt{\lambda}, \infty) \end{cases}$$ is concave and $\ge f(x)$. Now, the tangent line at $x=1$ for the graph $f_1$ is the tangent line for the graph of $f$. Therefore, the tangent line is above the graph of $f_1$, and therefore, above the graph of $f$.

$\bf{Added:}$ With all the preparations above, we can state:

if $a_1$, $\ldots$, $a_n$ are positive, $t_i \ge 0$, $\sum_{i=1}^n t_i = 1$, and $a\colon = \sum_{i=1}^n t_i a_i \le \sqrt{\lambda}$ then

$$\sum_{i=1}^n t_i f_{\lambda}(a_i) \le f_{\lambda}(a)$$