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

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

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

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

dau = xay ra khi a=b=c=1/3

trả lời 

=1/2

chúc bn

học tốt

Trả lời

Đáp số là 1/2

~ Học tốt ~

 minh moi hoc lop 5

ko bit đừng trả lời bừa nha mấy thánh ~~

chơi gunny đê mọi người ơi

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

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

Ta có: \(a+b+c=1\Leftrightarrow a^2+ab+ca=a\)

Thay vào ta có: \(\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{a^2+ab+ca+bc}}=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\)

Áp dụng Cauchy ngược: \(\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{a^2+ab+ca+bc}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}\)

Tương tự ta CM được: \(\sqrt{\frac{ab}{c+ab}}\le\frac{\frac{a}{c+a}+\frac{b}{c+b}}{2}\)

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

Cộng vế 3 BĐT trên ta được:

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

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

\(=\frac{1+1+1}{2}=\frac{3}{2}\)

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

Vậy \(Max_P=\frac{3}{2}\Leftrightarrow a=b=c=\frac{1}{3}\)

Ta có :

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

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

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

Áp dụng BĐT cauchy :

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

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

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

Cộng vế với vế :

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

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

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

có a+b+c=1 => c=(a+b+c).c=ac + bc + c²

=> c+ab=ac+bc+c²+ab=a(c+b)+c(b+c)=(c+a)(c+b)

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

tương tự có \(\hept{\begin{cases}a+bc=\left(a+b\right)\left(a+c\right)\\b+ca=\left(b+c\right)\left(b+a\right)\end{cases}\Rightarrow\sqrt{\frac{bc}{a+bc}}=\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{\frac{b}{a+b}+\frac{c}{a+c}}{2}}\)và \(\sqrt{\frac{ca}{b+ca}}=\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{\frac{c}{b+c}+\frac{a}{b+a}}{2}\)

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

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

vậy maxP=\(\frac{3}{2}\)đạt được khi a=b=c=\(\frac{1}{3}\)

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

thử dùng cô si đi

sửa ab thành a² mới làm như Thành được nhé :v

Ta có:

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

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

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

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

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

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

Cách 1:

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

Tương tự với \(\sqrt{\frac{bc}{a+bc}},\sqrt{\frac{ca}{b+ca}}\)rồi cộng các vế lại với nhau ta sẽ có

\(P\le\frac{3}{2}\)

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

Vậy....

Trả lời đi huhu

 ít người học lớp 9 lắm c

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

Tương tự: \(\sqrt{\frac{bc}{a+bc}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)\) ; \(\sqrt{\frac{ca}{b+ca}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{a+b}\right)\)

Cộng vế với vế: \(P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{b+c}+\frac{a}{a+c}+\frac{c}{a+c}\right)=\frac{3}{2}\)

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