Áp dụng BĐT svacxo
\(\dfrac{a^n}{\left(b+c\right)^n-a^n}+\dfrac{b^n}{\left(c+a\right)^n-b^n}+\dfrac{c^n}{\left(a+b\right)^n-c^n}\)
\(\dfrac{\left(a^n\right)^2}{\left(ab+ac\right)^n-\left(a^n\right)^2}+\dfrac{\left(b^n\right)^2}{\left(bc+ab\right)^n-\left(b^n\right)^2}+\dfrac{\left(c^n\right)^2}{\left(ac+bc\right)^n-\left(c^n\right)^2}\ge\dfrac{\left(\Sigma a^n\right)^2}{\left(ab+ac\right)^n+\left(ac+bc\right)^n+\left(bc+ab\right)^n-\Sigma\left(a^n\right)^2}\)
\(\ge\dfrac{3\Sigma\left(ab\right)^n}{\Sigma\left(ab+ac\right)^n+\left(ac+bc\right)^n+\left(bc+ab\right)^n-\Sigma\left(a^n\right)^2}\)
Do đó ta cần chứng minh
\(\dfrac{\Sigma\left(ab\right)^n}{\left(ab+ac\right)^n+\left(ac+bc\right)^n+\left(bc+ab\right)^n-\Sigma\left(a^n\right)^2}\ge\dfrac{1}{2^n-1}\)
\(\Leftrightarrow\left(2^n-1\right)\Sigma\left(ab\right)^n\ge\left(ab+ac\right)^n+\left(ac+bc\right)^n+\left(bc+ab\right)^n-\Sigma\left(a^n\right)^2\)
\(\Leftrightarrow\left(2^n-1\right)\Sigma\left(ab\right)^n+\Sigma\left(a^n\right)^2\ge\left(ab+ac\right)^n+\left(ac+bc\right)^n+\left(bc+ab\right)^n\)
Mặt khác \(\Sigma\left(a^n\right)^2\ge\Sigma\left(ab\right)^n\), do đó cần chứng minh
\(2^n\Sigma\left(ab\right)^n\ge\left(ab+ac\right)^n+\left(ac+bc\right)^n+\left(bc+ab\right)^n\)
\(\Leftrightarrow\Sigma\left(ab\right)^n\ge\left(\dfrac{ab+ac}{2}\right)^n+\left(\dfrac{ac+bc}{2}\right)^n+\left(\dfrac{bc+ab}{2}\right)^n\)
Cần chứng minh BĐT phụ sau :
\(\left(\forall a,b\ge0\right)x^{p+q}+y^{p+q}\ge\dfrac{1}{2}\left(x^p+y^p\right)\left(x^q+y^q\right)\)
\(\Leftrightarrow\left(x^p-y^p\right)\left(x^q-y^q\right)\ge0\) ( luôn đúng)
Áp dụng . \(\dfrac{x^n+y^n}{2}\ge\left(\dfrac{x+y}{2}\right)\left(\dfrac{x^{n-1}+y^{n-1}}{2}\right)....\ge\left(\dfrac{x+y}{2}\right)^n\)
Áp dụng .
\(\left(\dfrac{ab+ac}{2}\right)^n+\left(\dfrac{ac+bc}{2}\right)^n+\left(\dfrac{bc+ab}{2}\right)^n\le\left(\dfrac{\left(ab\right)^n+\left(ac\right)^n}{2}\right)+\left(\dfrac{\left(ac\right)^n+\left(bc\right)^n}{2}\right)+\left(\dfrac{\left(bc\right)^n+\left(ab\right)^n}{2}\right)=\Sigma\left(ab\right)^n\)
Vậy bài toán được chứng minh
Dấu "=" \(a=b=c\)