Question

[Trương Võ Thanh Ngân]

I wish the problem hadn't seem dificult for me

tìm lỗi sai và sửa

Answer 1

[Nguyễn Thị Ngọc Thơ]

I wish the problem hadn't seem => weren't dificult for me

Answer 2

[Nguyễn Việt Lâm]

\(P=\left(\frac{a}{a+b}\right)^4+\left(\frac{b}{b+c}\right)^4+\left(\frac{c}{c+a}\right)^4=\left(\frac{1}{1+\frac{b}{a}}\right)^4+\left(\frac{1}{1+\frac{c}{b}}\right)^4+\left(\frac{1}{1+\frac{a}{c}}\right)^4\)

Đặt \(\left\{{}\begin{matrix}\frac{b}{a}=x\\\frac{c}{b}=y\\\frac{a}{c}=z\end{matrix}\right.\) \(\Rightarrow xyz=1\)

\(P=\frac{1}{\left(1+x\right)^4}+\frac{1}{\left(1+y\right)^4}+\frac{1}{\left(1+z\right)^4}\)

Ta có \(\frac{1}{\left(1+x\right)^4}+\frac{1}{16}\ge\frac{1}{2\left(1+x\right)^2}\) tương tự với 2 biểu thức còn lại

\(\Rightarrow P+\frac{3}{16}\ge\frac{1}{2\left(1+x\right)^2}+\frac{1}{2\left(1+y\right)^2}+\frac{1}{2\left(1+z\right)^2}\)

\(\Rightarrow2P+\frac{3}{8}\ge\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}+\frac{1}{\left(1+z\right)^2}\)

Do \(xyz=1\Rightarrow\) trong 3 số luôn tồn tại ít nhất 1 số không nhỏ hơn 1, ta giả sử \(x\ge1\)

Áp dụng BĐT: \(\frac{1}{\left(1+y\right)^2}+\frac{1}{\left(1+z\right)^2}\ge\frac{1}{1+yz}\)

\(\Rightarrow2P+\frac{3}{8}\ge\frac{1}{1+yz}+\frac{1}{\left(1+x\right)^2}=\frac{x}{x+xyz}+\frac{1}{\left(1+x\right)^2}=\frac{x}{x+1}+\frac{1}{\left(1+x\right)^2}\)

\(\Rightarrow2P+\frac{3}{8}\ge\frac{x^2+x+1}{\left(x+1\right)^2}=\frac{4x^2+4x+4}{4\left(x+1\right)^2}=\frac{3\left(x+1\right)^2+\left(x-1\right)^2}{4\left(x+1\right)^2}=\frac{3}{4}+\frac{\left(x-1\right)^2}{4\left(x+1\right)^2}\ge\frac{3}{4}\)

\(\Rightarrow2P\ge\frac{3}{8}\Rightarrow P\ge\frac{3}{16}\Rightarrow P_{min}=\frac{3}{16}\) khi \(x=y=z=1\) hay \(a=b=c\)