Từ \(a^5+b^5=\left(a+b\right)\left(a^4-a^3b+a^2b^2-ab^3+b^4\right)\)

\(=\left(a+b\right)\left[a^2b^2+a^3\left(a-b\right)-b^3\left(a-b\right)\right]\)

\(=\left(a+b\right)\left[a^2b^2+\left(a-b\right)\left(a^3-b^3\right)\right]\)

\(=\left(a+b\right)\left[a^2b^2+\left(a-b\right)^2\left(a^2+ab+b^2\right)\right]\)

\(\ge\left(a+b\right)^2a^2b^2\forall a,b>0\)

\(\Rightarrow a^5+b^5+ab\ge ab\left[ab\left(a+b\right)+1\right]\)

\(\Rightarrow\frac{1}{a^5+b^5+ab}\le\frac{1}{ab\left(a+b\right)+1}=\frac{c}{a+b+c}\left(abc=1\right)\)

Tương tự cũng có: \(\frac{bc}{b^5+c^5+bc}\le\frac{a}{a+b+c};\frac{ca}{c^5+a^5+ca}\le\frac{b}{a+b+c}\)

Cộng theo vế ta có: 

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

Đẳng thức xảy ra khi \(a=b=c=1\)

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

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 a² + 1 \(\ge\)2a 

Khi đó \(\frac{1}{a^2+ab-a+5}=\frac{1}{a^2+1+ab-a+4}\le\frac{1}{2a+ab-a+4}=\frac{1}{ab+a+4}\)

Tương tự ta được \(\frac{1}{b^2+bc-b+5}\le\frac{1}{bc+b+4};\frac{1}{c^2+ac-c+5}\le\frac{1}{ac+c+4}\)

Cộng vế với vế => A \(\le\frac{1}{ab+a+4}+\frac{1}{bc+b+4}+\frac{1}{ca+c+4}\)

=> 4A \(\le\frac{4}{ab+a+1+3}+\frac{4}{bc+b+1+3}+\frac{4}{ca+c+1+3}\)

\(\le\frac{1}{ab+a+1}+\frac{1}{3}+\frac{1}{bc+b+1}+\frac{1}{3}+\frac{1}{ac+a+1}+\frac{1}{3}\)

\(=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+a+1}+1\)

\(=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{a^2bc+abc+ab}+1\)

\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+1=\frac{ab+a+1}{ab+a+1}+1=1+1=2\)

=> \(A\le\frac{1}{2}\)(Dấu "=" xảy ra <=> a = b = c = 1)

cho mik hỏi tí là làm sao ra được \(\frac{4}{ab+a+1+3}\le\frac{1}{ab+a+1}+\frac{1}{3}\) vậy ạ?

Dự đoán điểm rơi a = b = c = 1 

Ta có : \(\frac{1}{ab+a+1}+\frac{1}{3}\ge\frac{\left(1+1\right)^2}{ab+a+1+3}\)(BĐT Schwarz) 

\(=\frac{4}{a+b+c+4}\) (đpcm) 

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

Dễ dàng chứng minh được: 

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)với \(x,y>0\)(1)

Dấu bằng xảy ra \(\Leftrightarrow x=y>0\)

Ta có:

\(\frac{a}{bc\left(a+1\right)}=\frac{a}{abc+bc}=\frac{a}{ab+bc+ca+bc}=\frac{a}{\left(ab+bc\right)+\left(bc+ca\right)}\)

Áp dụng (1), ta được:

\(\frac{1}{ab+bc}+\frac{1}{bc+ca}\ge\frac{4}{\left(ab+bc\right)+\left(bc+ca\right)}\)

\(\Leftrightarrow\frac{1}{4\left(ab+bc\right)}+\frac{1}{4\left(bc+ca\right)}\ge\frac{1}{ab+bc+bc+ca}\)

\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{ab+bc+bc+ca}\)

\(\Leftrightarrow\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ca}\right)\ge\frac{a}{bc\left(a+1\right)}\left(2\right)\)

Dấu bằng xảy ra \(\Leftrightarrow b=c>0\)

Chúng minh tương tự, ta được:

\(\frac{b}{4}\left(\frac{1}{ab+ca}+\frac{1}{bc+ca}\right)\ge\frac{b}{ca\left(b+1\right)}\left(3\right)\)

Dấu bằng xảu ra \(\Leftrightarrow a=c>0\).

\(\frac{c}{4}\left(\frac{1}{ac+ab}+\frac{1}{ab+bc}\right)\ge\frac{c}{ab\left(c+1\right)}\left(4\right)\)

Từ (2), (3) và (4), ta được:

\(\frac{a}{bc\left(a+1\right)}+\frac{b}{ca\left(b+1\right)}+\frac{c}{ab\left(c+1\right)}\le\)\(\frac{a}{4}\left(\frac{1}{ab+bc}+\frac{1}{bc+ac}\right)+\frac{b}{4}\left(\frac{1}{ac+bc}+\frac{1}{ac+ab}\right)\)\(+\frac{c}{4}\left(\frac{1}{ab+bc}+\frac{1}{ab+ac}\right)\)

\(\Leftrightarrow P\le\frac{1}{4}.\left(\frac{a}{ab+bc}+\frac{c}{ab+bc}\right)+\frac{1}{4}\left(\frac{a}{bc+ac}+\frac{b}{bc+ac}\right)\)\(+\frac{1}{4}\left(\frac{b}{ab+ac}+\frac{c}{ab+ac}\right)\)

\(\Leftrightarrow P\le\frac{a+c}{4\left(ab+bc\right)}+\frac{a+b}{4\left(bc+ac\right)}+\frac{b+c}{4\left(ab+ac\right)}\)

\(\Leftrightarrow P\le\frac{a+c}{4b\left(a+c\right)}+\frac{a+b}{4c\left(a+b\right)}+\frac{b+c}{4a\left(b+c\right)}\)

\(\Leftrightarrow P\le\frac{1}{4b}+\frac{1}{4c}+\frac{1}{4a}\)

\(\Leftrightarrow P\le\frac{1}{4}\left(\frac{ab+bc+ca}{abc}\right)\)

\(\Leftrightarrow P\le\frac{1}{4}.\frac{abc}{abc}=\frac{1}{4}.1=\frac{1}{4}\)( vì \(ab+bc+ca=abc\))

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\ab+bc+ca=abc\end{cases}}\Leftrightarrow a=b=c=3\)

Vậy \(minP=\frac{1}{4}\Leftrightarrow a=b=c=3\)