tách ra mình làm cho. để cả đống này k làm đc đâu

ý a, áp dụng BĐT cô si có 

   a + b >= căn ab     dấu = xay ra a=b

b + c >= căn bc         dau = xay ra khi b=c

c+a >= căn ac           dau = xay ra khi a=c

công tung ve vao. rut gon ta dc điều phải chung minh

\(c+ab=\left(a+b+c\right)c+ab=ac+cb+c^2+ab=\left(a+c\right)\left(b+c\right)\)

Tương tự : \(a+bc=\left(a+b\right)\left(a+c\right);c+ab=\left(c+a\right)\left(c+b\right)\)

\(P=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)

áp dụng bất đẳng tức cauchy :

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

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

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

cộng vế theo vế 

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

\(\Leftrightarrow P\le\frac{1}{2}\left(\frac{a+c}{a+c}+\frac{b+c}{b+c}+\frac{a+b}{a+b}\right)=\frac{1}{2}\cdot3=\frac{3}{2}\)

dấu "=" xảy ra khi a=b=c=1/3

Có a+b+c=1 => c=(a+b+c).c=ac+bc+c²

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

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

Tương tự ta có \(\hept{\begin{cases}a+bc=\left(a+b\right)\left(a+c\right)\\b+ac=\left(b+a\right)\left(b+c\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\\\sqrt{\frac{ca}{b+ca}}=\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{\frac{c}{b+c}+\frac{a}{b+a}}{2}\end{cases}}}\)

\(\Rightarrow P\le\frac{\frac{b}{a+b}+\frac{c}{c+a}+\frac{c}{b+c}+\frac{a}{a+b}+\frac{a}{c+a}+\frac{b}{c+b}}{2}\)\(=\frac{\frac{a+c}{a+c}+\frac{c+b}{c+b}+\frac{a+b}{a+b}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

tham khảo

https://olm.vn/hoi-dap/detail/106887527253.html

\(VT=\sqrt{\frac{ab+2c^2}{a^2+ab+b^2}}+\sqrt{\frac{bc+2a^2}{b^2+bc+c^2}}+\sqrt{\frac{ca+2b^2}{c^2+ca+a^2}}\)

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

\(\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2c^2+2ab}+\frac{2\left(bc+2a^2\right)}{2a^2+b^2+c^2+2bc}+\frac{2\left(ca+2b^2\right)}{a^2+2b^2+c^2+2ca}\)

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

\(=2+ab+bc+ca=VP\) (Do a² + b² + c² = 1) => ĐPCM.

Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{\sqrt{3}}.\)

chăc là .............................. điền đi sẽ biếc a you ok ?

Làm đi làm lại nhiều rồi chán không muốn viết nữa vô TKHĐ xem hình ảnh

Ta có 

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

Tương tự, ta có

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

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

Cộng vế theo vế của 3 bđt ta được đpcm