Question

[zZz Cool Kid_new zZz]

Bất đẳng thức Netbitt'ss :3

Chứng minh rằng với mọi \(a,b,c,\alpha>0\) ta luôn có:

\(\frac{a^{\alpha}}{b+c}+\frac{b^{\alpha}}{c+a}+\frac{c^{\alpha}}{a+b}\ge\frac{3}{2}\cdot\frac{a^{\alpha}+b^{\alpha}+c^{\alpha}}{a+b+c}\)

Answer 1

[Tran Le Khanh Linh]

Không mất tính tổng quát giả sử \(a\ge b\ge c>0\Rightarrow\hept{\begin{cases}b+c\le a+c\le a+b\\\frac{a^a}{b+c}\ge\frac{b^a}{c+a}\ge\frac{c^a}{a+b}\end{cases}}\)

Sử dụng bất đẳng thức Chebyshev cho 2 dãy đơn ngược chiều ta có:

\(VT\left(1\right)=\frac{1}{2\left(a+b+c\right)}\left(\frac{a^a}{b+c}+\frac{b^a}{c+a}+\frac{c^a}{a+b}\right)\left[\left(b+c\right)+\left(c+a\right)+\left(a+b\right)\right]\ge\)

\(\frac{1}{2\left(a+b+c\right)}\cdot3\left[\frac{a^a}{b+c}\left(b+c\right)+\frac{b^a}{c+a}\left(c+a\right)+\frac{c^a}{a+b}\left(a+b\right)\right]=\frac{3\left(a^a+b^a+c^a\right)}{2\left(a+b+c\right)}\)\(=\frac{3}{2}\cdot\frac{a^a+b^a+c^a}{a+b+c}\)

=> đpcm

Answer 2

[Trí Tiên亗]

Ta có : \(\left(1+\sqrt{2019}\right)\sqrt{2020-2\sqrt{2019}}\)

\(=\left(1+\sqrt{2019}\right).\sqrt{2019-2\sqrt{2019}+1}\)

\(=\left(1+\sqrt{2019}\right)\sqrt{\left(\sqrt{2019}-1\right)^2}\)

\(=\left(1+\sqrt{2019}\right)\left(\sqrt{2019}-1\right)\)

\(=2019-1=2018\)