1)

   +)  Ta có

            \(\left(a-b\right)^2\ge0\)

       \(\Leftrightarrow a^2+b^2-2ab\ge0\)

        \(\Leftrightarrow a^2+b^2\ge2ab\)

        \(\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+b^2+2ab\)

        \(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

        \(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)  ( đpcm )

     + )   Theo phần trên

             \(a^2+b^2\ge2ab\)

           \(\Leftrightarrow a^2+b^2+2ab\ge4ab\)

           \(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

            \(\Leftrightarrow ab\le\frac{1}{4}\left(a+b\right)^2\)  ( đpcm )

2, 

Ta có: \(5\left(x^2+y^2+z^2\right)-9x\left(y+z\right)-18yz=0\Leftrightarrow5x^2-9x\left(y+z\right)+5\left(y+z\right)^2=28yz\le7\left(y+z\right)^2\)\(\Leftrightarrow5x^2-9x\left(y+z\right)-2\left(y+z\right)^2\le0\Leftrightarrow5\left(\frac{x}{y+z}\right)^2-9.\frac{x}{y+z}-2\le0\)\(\Leftrightarrow\left(5.\frac{x}{y+z}+1\right)\left(\frac{x}{y+z}-2\right)\le0\Leftrightarrow\frac{x}{y+z}\le2\)(Do \(5.\frac{x}{y+z}+1>0\forall x,y,z>0\))

\(\Rightarrow E=\frac{2x-y-z}{y+z}=2.\frac{x}{y+z}-1\le2.2-1=3\)

Đẳng thức xảy ra khi \(y=z=\frac{x}{4}\)

Lời giải:Vì $x^2+y^2+z^2=2$ nên:

$P=\frac{x^2+y^2+z^2}{x^2+y^2}+\frac{x^2+y^2+z^2}{y^2+z^2}+\frac{x^2+y^2+z^2}{z^2+x^2}-\frac{x^3+y^3+z^3}{2xyz}$

$=3+\frac{x^2}{y^2+z^2}+\frac{y^2}{x^2+z^2}+\frac{z^2}{x^2+y^2}-\frac{x^3+y^3+z^3}{2xyz}$

$\leq 3+\frac{x^2}{2yz}+\frac{y^2}{2xz}+\frac{z^2}{2xy}-\frac{x^3+y^3+z^3}{2xyz}$

(theo BĐT AM-GM)

$=3+\frac{x^3+y^3+z^3}{2xyz}-\frac{x^3+y^3+z^3}{2xyz}=3$

Vậy $P_{\max}=3$

Dấu "=" xảy ra khi $x=y=z=\sqrt{\frac{2}{3}}$

\(A=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\).Áp dụng BĐT Cauchy-Schwarz,ta có:

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

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

\(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)

Dấu "=" xảy ra khi x = y = z = 1/3

Vậy A min = 3/4 khi x=y=z=1/3

Bỏ chữ "Áp dụng bđt Cauchy-Schwarz,ta có:"giùm mình,nãy đánh nhầm ở bài làm trước mà quên xóa đi!

À mà để phải là tìm Max mới đúng chứ nhỉ?

Do đó,bạn sửa dòng: \(\ge3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\) đến hết thành:

"\(\le3-\frac{9}{\left(x+y+z\right)+\left(1+1+1\right)}=\frac{3}{4}\)

Dấu "=" xảy ra khi x=y=z=1/3

Vậy A max = 3/4 khi x=y=z=1/3

Bài 2 :

Ta có x , y , z là các số thực dương

Khi đó : \(5\left(x^2+y^2+z^2\right)-9x\left(y+z\right)-18yz=0\)

\(\Leftrightarrow5\frac{x^2}{\left(y+z\right)^2}+\frac{5\left(y^2+z^2\right)}{\left(y+z\right)^2}-\frac{9x}{y+z}-\frac{18yz}{\left(y+z\right)^2}=0\)

\(\Leftrightarrow5\left(\frac{x}{y+z}\right)^2-\frac{9x}{y+z}=\frac{18yz}{\left(y+z\right)^2}-\frac{5\left(y^2+z^2\right)}{\left(y+z\right)^2}\)

\(\le\frac{\frac{18\left(y+z\right)^2}{4}}{\left(y+z\right)^2}-\frac{\frac{5\left(y+z\right)^2}{2}}{\left(y+z\right)^2}=\frac{18}{4}-\frac{5}{2}=2\)

\(\Rightarrow5\left(\frac{x}{y+z}\right)^2-9.\frac{x}{y+z}\le2\)

Đặt \(\frac{x}{y+z}=a>0\) ta được : \(5a^2-9a-2\le0\)

\(\Leftrightarrow5a^2-10a+a-2\le0\Leftrightarrow\left(5a+1\right)\left(a-2\right)\le0\)

Dễ thấy :

\(5a+1>0\Rightarrow a-2\le0\Leftrightarrow a\le2\Leftrightarrow\frac{x}{y+z}\le2\)

Ta có :

\(Q=\frac{2x-y-z}{y+z}=\frac{2x}{y+z}-1\le2.2-1=3\)

Dấu " = '' xảy ra khi \(\left\{{}\begin{matrix}y=z\\\frac{x}{y+z}=2\end{matrix}\right.\) \(\Leftrightarrow x=4y=4z\)

Vậy GTLN của \(Q=3\Leftrightarrow x=4y=4z\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Rightarrow2\ge3x^2+2y^2+2z^2+y^2+z^2\) 

\(\Leftrightarrow2\ge3\left(x^2+y^2+z^2\right)\)

Có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\le2\)

\(\Rightarrow\)\(A^2\le2\) \(\Leftrightarrow A\in\left[-\sqrt{2};\sqrt{2}\right]\)

minA=-1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=-\sqrt{2}\\x=y=z\end{matrix}\right.\)  \(\Rightarrow x=y=z=-\dfrac{\sqrt{2}}{3}\)

maxA=1\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=\dfrac{\sqrt{2}}{3}\)

\(P\le\frac{1}{2}\left(\Sigma\frac{1}{\sqrt{xy}}\right)\le\frac{\left(xy+yz+zx\right)^2}{6x^2y^2z^2}\le\frac{\left(x^2+y^2+z^2\right)^2}{6x^2y^2z^2}=\frac{3}{2}\)

dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=1\)

mình nhầm :) làm lại nhé

\(P\le\frac{1}{2}\left(\Sigma\frac{1}{\sqrt{xy}}\right)\le\frac{\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}{6xyz}\le\frac{xy+yz+zx}{2xyz}\le\frac{x^2+y^2+z^2}{2xyz}=\frac{3}{2}\)