từ từ hồi trả lời cho câu này củng hơi khó cần thời gian suy nghĩ

Bài 1: \(T=\sqrt{\frac{x^3}{x^3+8y^3}}+\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\)

\(=\frac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\frac{2y^2}{\sqrt{y\left[y^3+\left(x+y\right)^3\right]}}\)

\(=\frac{x^2}{\sqrt{\left(x^2+2xy\right)\left(x^2-2xy+4y^2\right)}}+\frac{2y^2}{\sqrt{\left(xy+2y^2\right)\left(x^2+xy+y^2\right)}}\)

\(\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2y^2+\left(x+y\right)^2}\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2x^2+4y^2}=1\)

\(\Rightarrow T\ge1\)

Bài 2:

[Toán 10] Bất đẳng thức | Page 5 | HOCMAI Forum - Cộng đồng học sinh Việt Nam

Đặt vế trái là P

Ta có: \(P\ge\frac{x^2+1}{1+\frac{y^2+1}{2}+z^2}+\frac{y^2+1}{1+\frac{z^2+1}{2}+x^2}+\frac{z^2+1}{1+\frac{x^2+1}{2}+y^2}\)

Đặt \(\left(x^2+1;y^2+1;z^2+1\right)=\left(a;b;c\right)\Rightarrow a;b;c\ge1\)

\(P\ge\frac{2a}{b+2c}+\frac{2b}{c+2a}+\frac{2c}{a+2b}=2\left(\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}+\frac{c^2}{ca+2bc}\right)\)

\(P\ge\frac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\frac{6\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=2\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có : 

\(A=\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\)

\(\Rightarrow A=\frac{1+z+x^2}{1+y+z^2}+\frac{1+x+y^2}{1+z+x^2}+\frac{1+y+z^2}{1+x+y^2}\)

\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)

\(\Rightarrow A\ge3\sqrt[3]{\frac{1+z+x^2}{1+y+z^2}.\frac{1+x+y^2}{1+z+x^2}.\frac{1+y+z^2}{1+x+y^2}}\)

\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)

\(\Rightarrow A\ge3-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)

\(\Rightarrow A\ge3-\left(\frac{z}{y+2z}+\frac{x}{z+2x}+\frac{y}{x+2y}\right)\)

\(\Rightarrow A\ge3-\left(\frac{1}{2}-\frac{y}{2\left(y+2z\right)}+\frac{1}{2}-\frac{z}{2\left(z+2x\right)}+\frac{1}{2}-\frac{x}{2\left(x+2y\right)}\right)\)

\(\Rightarrow A\ge3-\frac{3}{2}+\frac{1}{2}\left(\frac{y}{y+2z}+\frac{z}{z+2x}+\frac{x}{x+2y}\right)\)

\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}+\frac{x^2}{x^2+2xy}\right)\)

\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{y^2+2yz+z^2+2xz+x^2+2xy}\right)\)

\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\right)\)

\(\Rightarrow A\ge2\)

Dấu " = " xảy ra khi \(x=y=z=1\)

Ta có : 

\(\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\)

\(\Rightarrow A=\frac{1+z+x^2}{1+y+z^2}+\frac{1+x+y^2}{1+z+x^2}+\frac{1+y+z^2}{1+x+y^2}\)

\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)

\(\Rightarrow A\ge3\sqrt[3]{\frac{1+z+x^2}{1+y+z^2}.\frac{1+x+y^2}{1+z+x^2}.\frac{1+y+z^2}{1+x+y^2}}\)

\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)

\(\Rightarrow A\ge3-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)

\(\Rightarrow A\ge3-\left(\frac{z}{y+2z}+\frac{x}{z+2x}+\frac{y}{x+2y}\right)\)

\(\Rightarrow A\ge3-\left(\frac{1}{2}-\frac{y}{2\left(y+2z\right)}+\frac{1}{2}-\frac{z}{2\left(z+2x\right)}+\frac{1}{2}-\frac{x}{2\left(x+2y\right)}\right)\)

\(\Rightarrow A\ge3-\frac{3}{2}+\frac{1}{2}\left(\frac{y}{y+2z}+\frac{z}{z+2x}+\frac{x}{x+2y}\right)\)

\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}+\frac{x^2}{x^2+2xy}\right)\)

\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{y^2+2yz+z^2+2xz+x^2+2xy}\right)\)

\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\right)\)

\(\Rightarrow A\ge2\)

Dấu " = " xảy ra khi x=y=z=1 

\(1+y+z^2\le1+\frac{1+y^2}{2}+z^2\)

\(\frac{1+x^2}{1+y+z^2}\ge\frac{2\left(1+x^2\right)}{1+b^2+2\left(1+c^2\right)}\)

Bất đẳng thức cần chứng minh tương đương

\(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\ge1\)

với \(a=1+x^2,b=1+y^2,c=1+z^2\)

\(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\ge\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge1\)

Chứng minh hoàn tất. Đẳng thức xảy ra khi \(a=b=c=1\)

\(A=\frac{x}{1+y^2}+\frac{y}{1+z^2}+\frac{z}{1+x^2}=x\left(1-\frac{y^2}{1+y^2}\right)+y\left(1-\frac{z^2}{1+z^2}\right)+z\left(1-\frac{x^2}{1+x^2}\right)\)

\(\Rightarrow A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)=\left(x+y+z\right)-\frac{xy+yz+zx}{2}\ge3-\frac{\frac{9}{3}}{2}=\frac{3}{2}\)

Dau '=' xay ra khi \(x=y=z=1\)

Vay \(A_{min}=\frac{3}{2}\)khi \(x=y=z=1\)