ĐK: \(\hept{\begin{cases}a\ge0\\b\ge0\\a+b\ge2011\end{cases}}\)

pt => \(a+b-2011=a+b+2011-2\sqrt{2011}\left(\sqrt{a}+\sqrt{b}\right)+2\left(\sqrt{ab}\right)\)

<=> \(2.2011-2\sqrt{2011}.\sqrt{a}-2\sqrt{2011}\sqrt{b}+2\sqrt{ab}=0\)

<=> \(\sqrt{2011}\left(\sqrt{2011}-\sqrt{a}\right)-\sqrt{b}\left(\sqrt{2011}-\sqrt{a}\right)=0\)

<=> \(\left(\sqrt{2011}-\sqrt{a}\right)\left(\sqrt{2011}-\sqrt{b}\right)=0\)

<=> a = 2011 và b = 2011 ( thỏa mãn đk )

Thử lại với phương trình ta thấy thỏa mãn

Vậy a= b = 2011.

Ta có: a + b + c = 2 nên \(2c+ab=c\left(a+b+c\right)+ab=ac+bc+c^2+ab\)

\(=\left(ca+c^2\right)+\left(bc+ab\right)=c\left(a+c\right)+b\left(a+c\right)\)\(=\left(b+c\right)\left(a+c\right)\)

Áp dụng BĐT Cô - si cho 2 số không âm:

\(\frac{1}{b+c}+\frac{1}{a+c}\ge2\sqrt{\frac{1}{\left(b+c\right)\left(a+c\right)}}\)(Vì a,b,c thực dương)

\(\Rightarrow\sqrt{\frac{1}{\left(b+c\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+c}\right)\)

\(\Rightarrow\frac{1}{\sqrt{2c+ab}}\le\frac{1}{2}\left(\frac{1}{b+c}+\frac{1}{a+c}\right)\)(cmt)

\(\Rightarrow\frac{ab}{\sqrt{ab+2c}}\le\frac{1}{2}\left(\frac{ab}{b+c}+\frac{ab}{a+c}\right)\)(nhân 2 vế cho ab thực dương)    (1)

(Dấu "="\(\Leftrightarrow\frac{1}{b+c}=\frac{1}{c+a}\Leftrightarrow b+c=c+a\Leftrightarrow a=b\))

Tương tự ta có: \(\frac{bc}{\sqrt{bc+2a}}\le\frac{1}{2}\left(\frac{bc}{b+a}+\frac{bc}{a+c}\right)\)(Dấu "="\(\Leftrightarrow b=c\))  (2)

\(\frac{ca}{\sqrt{ca+2b}}\le\frac{1}{2}\left(\frac{ca}{c+b}+\frac{ca}{b+a}\right)\)(Dấu "="\(\Leftrightarrow a=c\))  (3)

Cộng các BĐT (1) , (2) , (3), ta được:

\(P\le\frac{1}{2}\left(\frac{ab}{c+a}+\frac{ab}{c+b}+\frac{bc}{b+a}+\frac{cb}{c+a}+\frac{ac}{b+a}+\frac{ac}{c+b}\right)\)

\(\Rightarrow P\le\frac{1}{2}\left(\frac{b\left(c+a\right)}{c+a}+\frac{a\left(c+b\right)}{c+b}+\frac{c\left(b+a\right)}{b+a}\right)\)

\(\le\frac{1}{2}\left(a+b+c\right)=1\)

Vậy \(P=\frac{ab}{\sqrt{ab+2c}}\)\(+\frac{bc}{\sqrt{bc+2a}}\)\(+\frac{ca}{\sqrt{ca+2b}}\le1\)

(Dấu "="\(\Leftrightarrow a=b=c=\frac{2}{3}\))

Ta có:

\(\frac{ab}{\sqrt{ab+2c}}=\frac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\frac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{ab}{c+a}+\frac{ab}{c+b}\)

Tương tự:

\(\frac{bc}{\sqrt{bc+2a}}\le\frac{bc}{a+b}+\frac{bc}{a+c}\)

\(\frac{ca}{\sqrt{ca+2b}}\le\frac{ca}{b+c}+\frac{ca}{b+a}\)

Khi đó:

\(P\le\frac{ab}{a+c}+\frac{ab}{c+b}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ca}{b+c}+\frac{ca}{b+a}\)

\(=\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}+\frac{c\left(a+b\right)}{b+a}\)

\(=a+b+c=2\)

Dấu "=" xảy ra tại \(a=b=c=\frac{2}{3}\)

\(\sqrt{a^2-b^2}+\sqrt{2ab-b^2}>a\)

Có: \(\hept{\begin{cases}a^2-b^2>0\\2a-b^2>0\\a;b>0\end{cases}\Leftrightarrow a>b>0.}\)

 \(\sqrt{a^2-b^2}+\sqrt{2ab-b^2}>a\)(1)

<=> \(\sqrt{2ab-b^2}>a-\sqrt{a^2-b^2}\)

<=> \(2ab-b^2>a^2-2a\sqrt{a^2-b^2}+a^2-b^2\)

<=> \(b>a-\sqrt{a^2-b^2}\)

<=> \(a-b-\sqrt{a^2-b^2}< 0\)

<=> \(\sqrt{a-b}\left(\sqrt{a-b}-\sqrt{a+b}\right)< 0\)đúng vì \(\sqrt{a-b}-\sqrt{a+b}< 0\)

=>  (1) đúng.

Chia hai vế cho a, bất đẳng thức cần chứng minh được viết lại thành:

\(\sqrt{1-\left(\frac{b}{a}\right)^2}+\sqrt{2\left(\frac{b}{a}\right)-\left(\frac{b}{a}\right)^2}>1\)

Đặt \(\frac{b}{a}=x\Rightarrow0< x< 1\). Ta cần chứng minh:

\(\sqrt{1-x^2}+\sqrt{2x-x^2}>1\)

\(\Leftrightarrow2x-2x^2+2\sqrt{\left(1-x^2\right)\left(2x-x^2\right)}>0\) (bình phương 2 vế)

\(\Leftrightarrow2x\left(1-x\right)+2\sqrt{x\left(1-x\right)\left(1+x\right)\left(2-x\right)}>0\) (đúng)

Ta có đpcm.

Ta có:

\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-2+4-3\left(\frac{x}{y}+\frac{y}{x}\right)\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}+1\right)-3\left(\frac{x}{y}+\frac{y}{x}-1\right)\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}+2\right)\ge0\left(1\right)\)

Đến đây có 2 cách giải quyết

Cách 1:

\(\left(1\right)\Leftrightarrow\frac{x^2-xy+y^2}{xy}\cdot\frac{\left(x+y\right)^2}{xy}\ge0\)

\(\Leftrightarrow\frac{\left(x+y\right)^2\left(x^2-xy+y^2\right)}{x^2y^2}\ge0\)

\(\Leftrightarrow\frac{\left(x+y\right)^2\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]}{x^2y^2}\ge0\left(true!!!\right)\)

Cách 2 là đặt ẩn:)

Đặt \(\frac{x}{y}+\frac{y}{x}=t\Rightarrow t^2=\left(\frac{x}{y}+\frac{y}{x}\right)^2\ge4\cdot\frac{x}{y}\cdot\frac{y}{x}=4\)

\(\Rightarrow\left|t\right|\ge2\)

Khi đó ta có:

\(\left(t+1\right)\left(t-2\right)\ge0\)

Nếu \(t\ge2\Rightarrow t+1>0;t-2\ge0\Rightarrow\left(t+1\right)\left(t-2\right)\ge0\)

Nếu \(t\le-2\Rightarrow t+1< 0;t-2< 0\Rightarrow\left(t+1\right)\left(t-2\right)>0\)

=> đpcm

ĐK:3x+2>=0

  <=>3x>=-2

 <=> x>=-2/3

Trả lời:

Xét a = b = c = 1 thì thỏa mãn bài ra

Xét a ,b,c khác 1. do a,b,c có vai trò như nhau nên giả sử \(a\le b\le c\)

Áp dụng BĐT cô-si cho 3 số a+b+1,1-a,1-b, ta có :

\(\left(a+b+1\right)\left(1-a\right)\left(1-b\right)\le\left(\frac{a+b+1+1-a+1-b}{3}\right)^3=1\)

\(\Rightarrow\left(1-a\right)\left(1-b\right)\le\frac{1}{a+b+1}\)

\(\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\frac{1-c}{a+b+1}\)

Mà \(\frac{a}{b+c+1}\le\frac{a}{a+b+1};\frac{b}{a+c+1}\le\frac{b}{a+b+1}\)

\(\Rightarrow\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}\le\frac{a}{a+b+1}+\frac{b}{a+b+1}+\frac{c}{a+b+1}\)

do đó : \(\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

\(\le\frac{a+b+c}{a+b+1}+\frac{1-c}{a+b+1}=1\)

dấu " = " xảy ra khi a = b = c = 0

vậy ...

Có BĐT sau:

\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\Leftrightarrow2x^2+2y^2\ge x^2+2xy+y^2\Leftrightarrow\left(x-y\right)^2\ge0\left(true!\right)\)

Ta có:

\(x^8+y^8\ge\frac{\left(x^4+y^4\right)^2}{2}\ge\frac{\left[\frac{\left(x^2+y^2\right)^2}{2}\right]^2}{2}\ge2\)

Dấu "=" xảy ra tại a=b=1