Giải thích các bước giải:
Ta có:
$P=\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}\le\dfrac{1}{\sqrt{1+c^2}}$
$\to P=\dfrac{a}{ab+bc+ca+a^2}+\dfrac{b}{ab+bc+ca+b^2}$
$\to P=\dfrac{a}{(ab+bc)+(ca+a^2)}+\dfrac{b}{(ab+ac)+(bc+b^2)}$
$\to P=\dfrac{a}{b(a+c)+a(c+a)}+\dfrac{b}{a(b+c)+b(c+b)}$
$\to P=\dfrac{a}{(a+b)(a+c)}+\dfrac{b}{(b+a)(b+c)}$
$\to P=\dfrac{a(b+c)+b(a+c)}{(a+b)(b+c)(c+a)}$
$\to P=\dfrac{ab+ac+ab+bc}{(a+b)(b+c)(c+a)}$
$\to P=\dfrac{2ab+c(a+b)}{(a+b)(b+c)(c+a)}$
$\to P=\dfrac{\sqrt{ab}\cdot2\sqrt{ab}+c(a+b)}{(a+b)(b+c)(c+a)}$
$\to P\le\dfrac{\sqrt{ab}\cdot(a+b)+c(a+b)}{(a+b)(b+c)(c+a)}$
$\to P\le\dfrac{(\sqrt{ab}+c)\cdot(a+b)}{(a+b)(b+c)(c+a)}$
$\to P\le\dfrac{(\sqrt{a}\cdot\sqrt{b}+\sqrt{c}\cdot\sqrt{c})\cdot(a+b)}{(a+b)(b+c)(c+a)}$
$\to P\le\dfrac{\sqrt{(a+c)(b+c)}\cdot(a+b)}{(a+b)(b+c)(c+a)}$
$\to P\le\dfrac{1}{\sqrt{(a+c)(b+c)}}$
$\to P\le\dfrac{1}{\sqrt{ab+bc+ca+c^2}}$
$\to P\le\dfrac{1}{\sqrt{1+c^2}}$
