求证一个不等式

lemondian

设$a,b>0,n\inN^*,m=\dfrac{n(n-1)}{2}$，求证：$a^mb^m(a^n+b^n)\leqslant 2(\dfrac{a+b}{2})^{n^2}$。

Aluminiumor

只需证：
$$x^{\frac{n(n-1)}{2}}(x^n+1)<2\left(\frac{x+1}{2}\right)^{n^2},x>1$$
令
$$f(x)=n^2\ln\frac{x+1}{2}-\frac{n^2-n}{2}\ln x-\ln\left(\frac{x^n+1}{2}\right)$$
$$f'(x)=\frac{n}{2x(x+1)(x^n+1)}\left[(n-1)x^{n+1}-(n+1)x^n+(n-1)x-n+1\right]$$
令
$$g(x)=(n-1)x^{n+1}-(n+1)x^n+(n-1)x-n+1$$
$$g'(x)=(n^2-1)x^n-n(n+1)x^{n-1}+n-1$$
$$g''(x)=n(n^2-1)x^{n-2}(x-1)>0$$
$$\Longrightarrow g'(x)>g'(1)=0\Longrightarrow g(x)>g(1)=0\Longrightarrow f'(x)>0\Longrightarrow f(x)>f(1)=0$$
得证.