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 \(a=b=c=\frac{2}{3}\)

\(\frac{4ab}{1+ab}\le\frac{4ab}{2\sqrt{ab}}=2\sqrt{ab}\le a+b\)

Dấu "=" xảy ra khi a=b=1 

b: Tọa độ M là:

x=0 và y=1-3/2*0=1

Vì (d) đi qua M(0;1) và N(2;3) nên ta có hệ:

0a+b=1 và 2a+b=3

=>b=1; a=1

\(DK:x,y>0\)

HPT

\(\Leftrightarrow\hept{\begin{cases}y^2-3x^2y+2=0\left(1\right)\\x^2-3xy^2+2=0\left(2\right)\end{cases}}\)

\(\left(1\right)-\left(2\right)\Leftrightarrow-\left(x+y\right)\left(x-y\right)-3xy\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y+3xy\right)=0\)

\(\Leftrightarrow x=y\)

\(Q=\Sigma\frac{x^4}{x^2+\sqrt{xy.zx}}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+xy+yz+zx}\ge\frac{x^2+y^2+z^2}{2}\ge\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi x=y=z=1 

2ab =a^2+b^2+2ab

=>a^2+b^2=0

Mà a^2 >=0 ;b^2>=0 => a^2+b^2>=0 

=> a^2= b^2=0

=> a=b=0

Ta có : ab + ab = ( a + b )²

\(\Leftrightarrow\)( a + b )² - ( ab + ab )= 0

\(\Leftrightarrow\)a² + 2ab + b² - 2ab= 0

\(\Leftrightarrow\)( a² + b² ) + ( 2ab - 2ab ) = 0

\(\Leftrightarrow\)a² + b² = 0

\(\Leftrightarrow\)a² = 0 và b² = 0

\(\Leftrightarrow\)a = 0 và b = 0

Vậy : a = b = 0

Lời giải

Ta có: \(\left(a+b+\frac{1}{4}\right)^2=\frac{1}{16}\left(4a+4b-1\right)^2+\left(a+b\right)\ge a+b\)

Tương tự: \(\left(b+c+\frac{1}{4}\right)^2\ge b+c;\left(c+a+\frac{1}{4}\right)^2\ge c+a\)

Như vậy: \(L.H.S\left(VT\right)\ge\left(a+b\right)+\left(b+c\right)+\left(c+a\right)=\left(\frac{1}{\frac{1}{a}}+\frac{1}{\frac{1}{b}}\right)+\left(\frac{1}{\frac{1}{b}}+\frac{1}{\frac{1}{c}}\right)+\left(\frac{1}{\frac{1}{c}}+\frac{1}{\frac{1}{a}}\right)\)

\(\ge4\left(\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}+\frac{1}{\frac{1}{c}+\frac{1}{a}}\right)=R.H.S\left(VP\right)\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{8}\). Ta có đpcm.

khác cách tth xíu

Ta có:

\(VP=\Sigma_{cyc}\frac{4}{\frac{1}{a}+\frac{1}{b}}\le\Sigma_{cyc}\frac{4}{\frac{4}{a+b}}=2\left(a+b+c\right)\)

Gio ta di chung minh

\(VT\ge2\left(a+b+c\right)\)

Ta lai co:

\(VT=\Sigma_{cyc}\left(a+b+\frac{1}{4}\right)^2\ge\frac{\left[2\left(a+b+c\right)+\frac{3}{4}\right]^2}{3}\)

Chung minh

\(\frac{\left[2\left(a+b+c\right)+\frac{3}{4}\right]^2}{3}\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\left[2\left(a+b+c\right)-\frac{3}{4}\right]^2\ge0\) (đúng)

Dau '=' xay ra khi \(a=b=c=\frac{1}{8}\)