Question

Cho a;b là hai số dương thỏa mãn : \(a^2+b^2=6\) CM rằng \(\sqrt{3\left(a^2+6\right)}\) \(\geq\) \(\left(a+b\right)\sqrt{2}\)

Answer 1

Ta có: \(\sqrt{3\left(a^2+6\right)}\ge\left(a+b\right)\sqrt{2}\)

<=> \(3\left(a^2+6\right)\ge2\left(a+b\right)^2\)

<=> \(3\left(a^2+b^2+a^2\right)\ge2a^2+2b^2+4ab\)

<=> \(6a^2+3b^2\ge2a^2+2b^2+4ab\)

<=> \(4a^2-4ab+b^2\ge0\)

<=> \(\left(2a-b\right)^2\ge0\) ( Luôn đúng) => đpcm

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}2a=b\\a^2+b^2=6\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=\sqrt{\dfrac{6}{5}}=\dfrac{\sqrt{30}}{5}\\b=\dfrac{2\sqrt{30}}{5}\end{matrix}\right.\)