Xét bất đẳng thức phụ: \(\frac{19b^3-a^3}{ab+5b^2}\le4b-a\)(*)

Thật vậy: (*)\(\Leftrightarrow19b^3-a^3\le\left(4b-a\right)\left(ab+5b^2\right)=4ab^2-a^2b+20b^3-5ab^2\)\(\Leftrightarrow a^3+b^3-ab^2-a^2b\ge0\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)\(\Leftrightarrow\left(a^2-b^2\right)\left(a-b\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)*đúng với mọi a, b > 0*

Tương tự, ta được: \(\frac{19c^3-b^3}{cb+5c^2}\le4c-b\); \(\frac{19a^3-c^3}{ac+5a^2}\le4a-c\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{19b^3-a^3}{ab+5b^2}+\frac{19c^3-b^3}{cb+5c^2}+\frac{19a^3-c^3}{ac+5a^2}\le3\left(a+b+c\right)\)

Đẳng thức xảy ra khi  a = b = c

\(DK:0\le x\le1\)

pt tren td voi:

\(3.\left[\frac{x-\left(1-x\right)}{\sqrt{x}-\sqrt{1-x}}\right]=3+2\sqrt{x}.\sqrt{1-x}\)

\(\Leftrightarrow3.\left(\sqrt{x}+\sqrt{1-x}\right)=3+2.\sqrt{x}.\sqrt{1-x}\)(1)

Dat \(\hept{\begin{cases}\sqrt{x}=a\left(a\ge0\right)\\\sqrt{1-x}=b\left(b\ge0\right)\end{cases}}\)=> a^2+b^2=1

tu (1), ta co pt: 3(a+b)=3+2ab

bp 2 ve: 9.(a+b)^2=(3+2ab)

=> 6ab-4a^2b^2=0 ( a^2+b^2=1)

=> ....

ta sẽ chứng minh \(\frac{x^{2020}+y^{2020}}{x^{2019}+y^{2019}}\ge\frac{x+y}{2}\Leftrightarrow2\left(x^{2020}+y^{2020}\right)\ge\left(x+y\right)\left(x^{2019}+y^{2019}\right)\)

\(\Leftrightarrow x^{2020}+y^{2020}\ge xy^{2019}+x^{2019}y\)(*)    điều này đúng do

\(\hept{\begin{cases}2019x^{2020}+y^{2020}=x^{2020}+..+x^{2020}+y^{2020}\ge2020.x^{2019}y\\x^{2020}+2019y^{2020}=x^{2020}+y^{2020}+..+y^{2020}\ge2020.xy^{2019}\end{cases}}\)cộng hai BDT lại ta sẽ có (*) đúng

vậy \(\frac{x^{2020}+y^{2020}}{x^{2019}+y^{2019}}+\frac{y^{2020}+z^{2020}}{y^{2019}+z^{2019}}+\frac{z^{2020}+x^{2020}}{z^{2019}+x^{2019}}\ge\frac{x+y}{2}+\frac{y+z}{2}+\frac{x+z}{2}=2020\)

dấu bằng xảy ra khi \(x=y=z=\frac{2020}{3}\)