Inequality.such as Nesbitt

Question

Let $a,b,c >0 $ , prove that:

$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} \leq \dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}$$

Answer 1

Solution 1: \begin{align*}&\sum\dfrac{a^2}{b^2+c^2}-\sum\dfrac{a}{b+c}=\sum_{cyc}\dfrac{ab(a-b)+ac(a-c)}{(b^2+c^2)(b+c)}\\ &=\sum_{cyc}\dfrac{ab}{(b^2+c^2)(b+c)}(a-b)+\dfrac{ac}{(b^2+c^2)(b+c)}(a-c)\\ &=\sum_{cyc}\left[\dfrac{ab}{(b^2+c^2)(b+c)}-\dfrac{ab}{(c+a)(c^2+a^2)}\right](a-b)\\ &=\sum_{cyc}\dfrac{(a+c)(a^2+c^2)-(b+c)(b^2+c^2)}{(a+c)(b+c)(a^2+c^2)(b^2+c^2)}(a-b)ab\\ &=\sum_{cyc}\dfrac{(a-b)(a^2+b^2+c^2+ab+bc+ac)}{(a+c)(b+c)(a^2+c^2)(b^2+c^2)}(a-b)ab\\ &=\sum_{cyc}\dfrac{ab(a-b)^2(a^2+b^2+c^2+ab+bc+ac)}{(a+c)(b+c)(b^2+c^2)(a^2+c^2)}\\ &\ge 0 \end{align*}

Solution 2:

let $$f(x)=\dfrac{a^x}{b^x+c^x}+\dfrac{b^x}{c^x+a^x}+\dfrac{c^x}{a^x+b^x}$$ then we have \begin{align*} f'(x)&=\sum_{cyc}\dfrac{a^x(b^x+c^x)\ln{a}-a^x(b^x\ln{b}+c^x\ln{c})}{(b^x+c^x)^2}\\ &=\sum_{cyc}\dfrac{a^xb^x(a^x-b^x)(\ln{a}-\ln{b})(2c^x+a^x+b^x)}{(a^x+b^x)^2(b^x+c^x)^2}\\ &\ge 0 \end{align*}

$$f(2)\ge f(1)$$ By done!