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từ từ hồi trả lời cho câu này củng hơi khó cần thời gian suy nghĩ
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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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 \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{x}{1+y^2}+\frac{y}{1+z^2}+\frac{z}{1+x^2}=x\left(1-\frac{y^2}{1+y^2}\right)+y\left(1-\frac{z^2}{1+z^2}\right)+z\left(1-\frac{x^2}{1+x^2}\right)\)
\(\Rightarrow A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)=\left(x+y+z\right)-\frac{xy+yz+zx}{2}\ge3-\frac{\frac{9}{3}}{2}=\frac{3}{2}\)
Dau '=' xay ra khi \(x=y=z=1\)
Vay \(A_{min}=\frac{3}{2}\)khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1,theo giả thiết => \(x^2+y^2+z^2=x+y+z\)
mà \(3\left(x^2+y^2+z^2\right)>=\left(x+y+z\right)^2\)(bunhiacopxki)
=>\(x+y+z=< 3\)
ta có:\(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}>=\frac{9}{x+y+z+6}=1\)(cauchy schwarz)
p= 1+2 : 1 + 3 x 2 +1 + 2 : 1 + 3 + 4 + 1 +2 : 1 + 2 + 3
= 30