\(x+y\le xy\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\le1\)

\(M=\dfrac{1}{2\left(x^2+y^2\right)+y^2}+\dfrac{1}{2\left(x^2+y^2\right)+x^2}\le\dfrac{1}{4xy+y^2}+\dfrac{1}{4xy+x^2}\)

\(B\le\dfrac{1}{25}\left(\dfrac{4}{xy}+\dfrac{1}{y^2}\right)+\dfrac{1}{25}\left(\dfrac{4}{xy}+\dfrac{1}{x^2}\right)=\dfrac{1}{25}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}+\dfrac{6}{xy}\right)\)

\(M\le\dfrac{1}{25}\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2+\dfrac{3}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\right]=\dfrac{1}{10}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le\dfrac{1}{10}\)

\(M_{max}=\dfrac{1}{10}\) khi \(x=y=2\)

\(x^{2011}+x^{2011}+1+...+1\) (2009 số 1) \(\ge2011\sqrt[2011]{x^{4022}}=2011x^2\)

Tương tự:

\(2y^{2011}+2009\ge2011y^2\); \(2z^{2011}+2009\ge2011z^2\)

Cộng vế:

\(2\left(x^{2011}+y^{2011}+z^{2011}\right)+6027\ge2011\left(x^2+y^2+z^2\right)\)

\(\Rightarrow2011\left(x^2+y^2+z^2\right)\le6033\)

\(\Rightarrow x^2+y^2+z^2\le3\)

\(4=2^x+2^y\ge2\sqrt{2^{x+y}}\Rightarrow2^{x+y}\le4\Rightarrow x+y\le2\)

\(\Rightarrow xy\le1\)

\(P=4x^2y^2+2x^3+2y^3+10xy\)

\(P=4x^2y^2+10xy+2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

\(P\le4x^2y^2+10xy+4\left(4-3xy\right)=4x^2y^2-2xy+16\)

Đặt \(xy=t\Rightarrow0< t\le1\)

Xét hàm \(f\left(t\right)=4t^2-2t+16\) trên \((0;1]\)

\(\Rightarrow...\)

Đáp án C.

Ta có:

G T ⇔ 5 x + 2 y + x + 2 y − 3 − x − 2 y = 5 x y − 1 − 3 1 − x y + x y − 1.

Xét hàm số

f t = 5 t + t − 3 − t ⇒ f t = 5 t ln 5 + 1 + 3 − t ln 3 > 0   ∀ t ∈ ℝ

Do đó hàm số đồng biến trên ℝ  suy ra f x + 2 y = f x y − 1 ⇔ x + 2 y = x y − 1

⇔ x = 2 y + 1 y − 1 ⇒ T = 2 y + 1 y − 1 + y . Do x > 0 ⇒ y > 1  

Ta có:  T = 2 + y + 3 y − 1 = 3 + y − 1 + 3 y − 1 ≥ 3 + 2 3 .

Đáp án C.

Ta có: GT

<=> 5^x+2y + x + 2y – 3^–x–2y = 5^xy–1 – 3^1–xy + xy – 1.

X é t   h à m   s ố   f t = 5 t + t - 3 - t

⇒ f t = 5 t ln 5 + 1 + 3 - t ln 3 > 0   ∀ t ∈ ℝ

Do đó hàm số đồng biến trên  ℝ suy ra

f(x+2y) = f(xy – 1) <=> x+ 2y = xy – 1

⇔ x = 2 y + 1 y - 1 ⇒ T = 2 y + 1 y - 1 + y .

Do x > 0 => y > 1.

Ta có:

T = 2 + y + 3 y - 1 = 3 + y - 1 + 3 y - 1 ≥ 3 + 2 3 .

a) \(\left\{{}\begin{matrix}a=x\\b=2y\\c=3z\end{matrix}\right.\Rightarrow a+b+c=2;a,b,c>0\)

\(\Rightarrow S=\sqrt{\dfrac{\dfrac{ab}{2}}{\dfrac{ab}{2}+c}}+\sqrt{\dfrac{\dfrac{bc}{2}}{\dfrac{bc}{2}+a}}+\sqrt{\dfrac{ca}{ca+2b}}\)

\(=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)

Vì a,b,c>0 nên áp dụng BĐT AM-GM, ta có: 

 \(\sqrt{\dfrac{ab}{ab+2c}}=\sqrt{\dfrac{ab}{ab+\left(a+b+c\right)c}}=\sqrt{\dfrac{ab}{c^2+bc+ca+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\)

\(=\sqrt{\dfrac{a}{a+c}}.\sqrt{\dfrac{b}{b+c}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\) 

\(\sqrt{\dfrac{bc}{bc+2a}}=\sqrt{\dfrac{bc}{\left(b+a\right)\left(c+a\right)}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\)

\(\sqrt{\dfrac{ca}{ca+2b}}=\sqrt{\dfrac{ca}{\left(c+b\right)\left(a+b\right)}}\le\dfrac{1}{2}\left(\dfrac{c}{b+c}+\dfrac{a}{a+b}\right)\)

\(\Rightarrow S\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi và chỉ khi: a=b=c=2/3=>\(\left(x,y,z\right)=\left\{\dfrac{2}{3};\dfrac{1}{3};\dfrac{2}{9}\right\}\)

Cho \(xy=1\)và \(x,y>0\)

Tìm \(M_{max}=\frac{x}{x^4+y^2}+\frac{y}{x^2+y^4}\)

\(M=\frac{x}{x^4+\frac{1}{x^2}}+\frac{x}{y^2+\frac{1}{y^2}}\)

\(M=\frac{x^4}{x^6+1}+\frac{y^3}{y^6+1}\)

Áp dụng BĐT Cauchy

\(x^6+1\ge2x^3=>\frac{x^2}{x^6+1}\le\frac{1}{2}\)

Tương tự \(\frac{y^3}{y^6+1}\le\frac{1}{2}\)

\(=>M\le1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}xy=1\\x=1\\y=1\end{cases}}\Leftrightarrow x=y=1\)

Vậy \(M_{max}=1\)khi \(x=y=1\)

Đáp án là C