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Ta có \(\left(2x^2+y^2+3\right)\left(2+1+3\right)\ge\left(2x+y+3\right)^2\)

=> \(\frac{1}{\sqrt{2x^2+y^2+3}}\le\frac{\sqrt{6}}{2x+y+3}\)

Mà \(\frac{1}{2x+y+3}=\frac{1}{x+x+y+1+1+1}\le\frac{1}{36}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+3\right)\)

=> \(\frac{1}{\sqrt{2x^2+y^2+3}}\le\frac{\sqrt{6}}{36}\left(\frac{2}{x}+\frac{1}{y}+3\right)\)

Khi đó 

\(P\le\frac{\sqrt{6}}{36}\left(\frac{3}{x}+\frac{3}{y}+\frac{3}{z}+9\right)=\frac{\sqrt{6}}{36}.18=\frac{\sqrt{6}}{2}\)

Dấu bằng xảy ra khi x=y=z=1

Vậy \(MaxP=\frac{\sqrt{6}}{2}\)khi x=y=z=1

dễ vãi mà ko giải đc NGU

\(3-P=1-\frac{x}{x+1}+1-\frac{y}{y+1}+1-\frac{z}{z+1}\)

\(=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{9}{x+y+z+3}=\frac{9}{1+3}=\frac{9}{4}\)

\(\Rightarrow P\le\frac{3}{4}\)

Dấu "=" xảy ra tại \(x=y=z=\frac{1}{3}\)

2/\(LHS\ge\frac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)

\(\ge\frac{\left(a+b+c\right)^2}{a+b+c+\frac{1+b+c}{3}+\frac{1+c+a}{3}+\frac{1+a+b}{3}}=\frac{3}{2}\)

Bài 1 : 

\(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)

Thay \(x+y+z=1\)vào biểu thức 

\(\Rightarrow P=\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\)

Áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\forall a,b>0\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2x+y+z}=\frac{x}{x+y+x+z}\le\frac{x}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{y}{x+2y+z}=\frac{y}{x+y+y+z}\le\frac{y}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{z}{x+y+2z}=\frac{z}{x+z+y+z}\le\frac{z}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)

\(\Rightarrow VT\le\frac{x}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)+\frac{y}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)

\(+\frac{z}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

\(\Rightarrow VT\le\frac{x}{4\left(x+y\right)}+\frac{y}{4\left(x+y\right)}+\frac{x}{4\left(x+z\right)}+\frac{z}{4\left(x+z\right)}+\frac{y}{4\left(y+z\right)}\)

\(+\frac{z}{4\left(y+z\right)}\)

\(\Rightarrow VT\le\frac{x+y}{4\left(x+y\right)}+\frac{x+z}{4\left(x+z\right)}+\frac{y+z}{4\left(y+z\right)}\)

\(\Rightarrow VT\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)

\(\Rightarrow P\le\frac{3}{4}\)

Vậy \(P_{max}=\frac{3}{4}\)

Dấu " = " xảy ra khi \(x=y=z\)

Chúc bạn học tốt !!!

helloo

Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)

Khi đó BĐT <=>

 \(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)

<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)

<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)

<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)

Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)

<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)

<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng

Khi đó (1) <=> 

\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\) 

<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)

Áp dụng buniacopxki cho vế phải ta có 

\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)

                                                                                                       \(=\sqrt{2\left(x+y+z\right)}\)

=> BĐT được CM

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

Áp dụng bất đẳng thức Minkowski ta có:

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

\(\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{9}{x+y+z}\right)^2}=\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)

\(=\sqrt{\left[\left(x+y+z\right)^2+\frac{1}{\left(x+y+z\right)^2}\right]+\frac{80}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{2\sqrt{\left(x+y+z\right)^2\cdot\frac{1}{\left(x+y+z\right)^2}}+\frac{80}{1}}=\sqrt{82}\)

Dấu "=" xảy ra khi: \(x=y=z=\frac{1}{3}\)

Áp dụng bất đẳng thức Minkowski ta có:

√x2+1x2 +√y2+1y2 +√z2+1z2 ≥√(x+y+z)2+(1x +1y +1z )2

≥√(x+y+z)2+(9x+y+z )2=√(x+y+z)2+81(x+y+z)2 

=√[(x+y+z)2+1(x+y+z)2 ]+80(x+y+z)2 

≥√2√(x+y+z)2·1(x+y+z)2 +801 =√82

Dấu "=" xảy ra khi: x=y=z=13 

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

Ta có: \(x+y+z=xyz\Rightarrow x=\frac{x+y+z}{yz}\Rightarrow x^2=\frac{x^2+xy+xz}{yz}\Rightarrow x^2+1=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)\(\Rightarrow\sqrt{x^2+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}\le\frac{2+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

Tương tự: \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3.\frac{xy+yz+zx}{xyz}\)\(\le3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}=\frac{\left(xyz\right)^2}{xyz}=xyz\)

Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)

a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)

b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)

\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)