a,b,c>0,ab+ac+bc=1 chứng minh rằng P=a/(1+a^2) +b/(1+b^2)<= 1/căn(1+c^2)

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1. Giải thích các bước giải:

    Ta có:

   $P=\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}\le\dfrac{1}{\sqrt{1+c^2}}$

   $\to P=\dfrac{a}{ab+bc+ca+a^2}+\dfrac{b}{ab+bc+ca+b^2}$

   $\to P=\dfrac{a}{(ab+bc)+(ca+a^2)}+\dfrac{b}{(ab+ac)+(bc+b^2)}$

   $\to P=\dfrac{a}{b(a+c)+a(c+a)}+\dfrac{b}{a(b+c)+b(c+b)}$

   $\to P=\dfrac{a}{(a+b)(a+c)}+\dfrac{b}{(b+a)(b+c)}$

   $\to P=\dfrac{a(b+c)+b(a+c)}{(a+b)(b+c)(c+a)}$

   $\to P=\dfrac{ab+ac+ab+bc}{(a+b)(b+c)(c+a)}$

   $\to P=\dfrac{2ab+c(a+b)}{(a+b)(b+c)(c+a)}$

   $\to P=\dfrac{\sqrt{ab}\cdot2\sqrt{ab}+c(a+b)}{(a+b)(b+c)(c+a)}$

   $\to P\le\dfrac{\sqrt{ab}\cdot(a+b)+c(a+b)}{(a+b)(b+c)(c+a)}$

   $\to P\le\dfrac{(\sqrt{ab}+c)\cdot(a+b)}{(a+b)(b+c)(c+a)}$

   $\to P\le\dfrac{(\sqrt{a}\cdot\sqrt{b}+\sqrt{c}\cdot\sqrt{c})\cdot(a+b)}{(a+b)(b+c)(c+a)}$

   $\to P\le\dfrac{\sqrt{(a+c)(b+c)}\cdot(a+b)}{(a+b)(b+c)(c+a)}$

   $\to P\le\dfrac{1}{\sqrt{(a+c)(b+c)}}$

   $\to P\le\dfrac{1}{\sqrt{ab+bc+ca+c^2}}$

   $\to P\le\dfrac{1}{\sqrt{1+c^2}}$

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