Ta có \(\dfrac{a^2+b^2}{a+b}\ge\sqrt[3]{\dfrac{a^3+b^3}{2}}\Leftrightarrow\left(\dfrac{a^2+b^2}{a+b}\right)^3\ge\dfrac{a^3+b^3}{2}\Leftrightarrow2\left(a^2+b^2\right)^3\ge\left(a^3+b^3\right)\left(a+b\right)^3\Leftrightarrow2a^6+6a^4b^2+6a^2b^4+2b^6\ge\left(a^3+b^3\right)\left(a+b\right)^3\Leftrightarrow a^6+b^6-2a^3b^3+3a^4b^2+3a^2b^4-3ab^5-3a^5b\ge0\Leftrightarrow\left(a^3-b^3\right)^2+\left(b-a\right)3ab\left(a^3-b^3\right)\ge0\Leftrightarrow\left(a^3-b^3\right)\left(a^3-b^3+3ab\left(b-a\right)\right)\ge0\Leftrightarrow\left(a-b\right)^4\left(a^2+ab+b^2\right)\ge0\)(luôn đúng)
Ta có \(\sqrt{\dfrac{a^2+b^2}{2}}\ge\dfrac{a+b}{2}\Leftrightarrow\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\Leftrightarrow2a^2+2b^2\Leftrightarrow\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
BĐT \(\dfrac{a+b}{2}\ge\sqrt{ab}\) là BĐT Cauchy, bạn tự chứng minh nhá.
Ta có \(a+b\ge2\sqrt{ab}\Leftrightarrow\dfrac{2ab}{a+b}\le\dfrac{2ab}{2\sqrt{ab}}=\sqrt{ab}\)
Dấu = xảy ra <=> a = b.
