Ta có:

D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18

D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18

D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1

D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1

Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3

Hay x = 5 , y = -3

Đc chx bạn

Do \(x;y\in\left[0;2\right]\Rightarrow\left\{{}\begin{matrix}x\left(2-x\right)\ge0\\y\left(2-y\right)\ge0\end{matrix}\right.\) \(\Rightarrow2x^2+4y^2\le4x+8y\)

\(P\le3^0+5^0+3^z+4\left(x+2y\right)=2+3^z+4\left(6-z\right)=3^z-4z+26\)

Xét hàm \(f\left(z\right)=3^z-4z+26\) trên \(\left[0;2\right]\)

\(f'\left(z\right)=3^z.ln3-4=0\Rightarrow z=log_3\left(\dfrac{4}{ln3}\right)=a\)

\(f\left(0\right)=27\) ; \(f\left(2\right)=27\); \(f\left(a\right)\approx-1,1\)

\(\Rightarrow f\left(z\right)\le27\Rightarrow maxP=27\)

(Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(0;2;2\right)\))

Ồ mà khoan, bài trước bị nhầm lẫn ở chỗ \(3^{2x-x^2}+5^{2y-y^2}\ge3^0+5^0\) mới đúng, ko để ý bị ngược dấu đoạn này

Vậy giải cách khác:

\(0\le x;y;z\le2\Rightarrow x\left(2-x\right)\ge0\Rightarrow2x-x^2\ge0\)

Lại có: \(2x-x^2=1-\left(x-1\right)^2\le1\)

\(\Rightarrow0\le2x-x^2\le1\)

Tương tự ta có: \(0\le2y-y^2\le1\)

Xét hàm: \(f\left(t\right)=3^t-2t\) trên \(\left[0;1\right]\)

\(f'\left(t\right)=3^t.ln3-2=0\Rightarrow t=log_3\left(\dfrac{2}{ln3}\right)=a\)

\(f\left(0\right)=1;\) \(f\left(1\right)=1\) ; \(f\left(a\right)\approx0,73\)

\(\Rightarrow f\left(t\right)\le1\Rightarrow3^t-2t\le1\Rightarrow3^t\le2t+1\)

\(\Rightarrow3^{2x-x^2}\le2\left(2x-x^2\right)+1\)

Hoàn toàn tương tự, ta chứng minh được: 

\(5^t\le4t+1\) với \(t\in\left[0;1\right]\Rightarrow5^{2y-y^2}\le4\left(2y-y^2\right)+1\)

\(3^t\le4t+1\) với \(t\in\left[0;2\right]\Rightarrow3^z\le4z+1\)

\(\Rightarrow P\le2\left(2x-x^2\right)+4\left(2y-y^2\right)+4z+3+2x^2+4y^2=4\left(x+2y+z\right)+3=27\)

Lần này thì ko sai được rồi

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

\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)

\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)

Dấu \("="\Leftrightarrow x=y=z=1\)

\(\sqrt{5x^2+2xy+2y^2}=\sqrt{4x^2+2xy+y^2+x^2+y^2}\ge\sqrt{4x^2+2xy+y^2+2xy}=2x+y\)

\(\Rightarrow\dfrac{1}{\sqrt{5x^2+2xy+2y^2}}\le\dfrac{1}{2x+y}=\dfrac{1}{x+x+y}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{1}{9}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)

Tương tự:

\(\dfrac{1}{\sqrt{5y^2+2yz+2z^2}}\le\dfrac{1}{9}\left(\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{\sqrt{5z^2+2zx+2x^2}}\le\dfrac{1}{9}\left(\dfrac{2}{z}+\dfrac{1}{x}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=1\)

\(P_{max}=1\) khi \(x=y=z=1\)

Ta có: 

\(2x^2+xy+2y^2=x^2+y^2+\frac{3}{4}\left(x+y\right)^2+\frac{1}{4}\left(x-y\right)^2\)

\(\ge\frac{2\left(x+y\right)^2}{4}+\frac{3\left(x+y\right)^2}{4}=\frac{5\left(x+y\right)^2}{4}\)

\(\Rightarrow\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}\left(x+y\right)\). Tương tự ta có:

\(\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right);\sqrt{2z^2+xz+2x^2}\ge\frac{\sqrt{5}}{2}\left(x+z\right)\)

\(\Rightarrow M\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(x+z\right)\)

\(=\sqrt{5}\left(x+y+z\right)=\sqrt{5}\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)

Cho mình hối tại sao đẳng thức sảy ra x=y=z=1/3 vậy

Dấu = khi \(\hept{\begin{cases}x=y=z\\x+y+z=1\end{cases}\Rightarrow}x=y=z=\frac{1}{3}\)