Bài 2:

\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)

\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)

\(\Rightarrow P\ge\sqrt[3]{3}\)

Dấu bằng xẩy ra khi a=b=c=3

Bài 1: 

 \(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)

Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)

\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)

Áp dụng bđt AM-GM ta có:

 \(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)

\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)

\(\Rightarrow\)(*) luôn đúng

Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)

Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)

Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)

\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)

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Làm sai rồi ạ

Nếu bài toán ko yêu cầu a, b, c, d >= 1:

\(4=ab+bc+cd+da=\left(a+c\right)\left(b+d\right)\le\frac{\left(a+c+b+d\right)^2}{4}\)

\(\Rightarrow\left(a+b+c+d\right)^2\ge16\Rightarrow a+b+c+d\ge4\)

\(\frac{a^4}{a^3+2b^3}=\frac{a\left(a^3+2b^3\right)-2ab^3}{a^3+2b^3}=a-\frac{2ab^3}{a^3+b^3+b^3}\ge a-\frac{2ab^3}{3\sqrt[3]{a^3.b^3.b^3}}=a-\frac{2}{3}b\)

Tương tự với các cụm còn lại, công theo vế và áp dụng \(a+b+c+d\ge4\), ta được đpcm.

\(a;b;c;d\ge1\Rightarrow ab+bc+cd+da\ge4\)

Dấu bằng chỉ xảy ra khi mổi số bằng 1

=)) đề có vẻ sai sai bác lazy ơi 

Ta có:

\(\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge\left(a+b+c+d\right).\frac{16}{\left(a+b+c+d\right)}=16\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge4\)

Dấu = xảy ra khi \(a=b=c=d=1\)

Qúa khó

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Thế mà còn xưng là KUDO SHINICHI

mới lớp 6 ak

\(\frac{d}{b^2}\) hay \(\frac{b^2}{d}\)hả bạn?

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

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

Do đó: \(VT=\frac{a^2}{c}+\frac{b}{d^2}=\frac{d^2}{b}+\frac{b}{d^2}\ge2\sqrt{\frac{d^2}{b}.\frac{b}{d^2}}=2\)

MIN=1=>a=b=c=1

ta có 

\(\frac{a}{1+2b^3}=\frac{a\left(1+2b^3\right)-2ab^3}{1+2b^3}=a-\frac{2ab^3}{1+2b^3}\)

Vì \(1+2b^3\ge3b^2\left(cosi\right)\)

\(\Rightarrow a-\frac{2ab^3}{a+2b^3}\ge a-\frac{2}{3}ab\)

cmtt ta đc 

P\(\ge a+b+c-\frac{2}{3}\left(ab+bc+ca\right)\)

\(P\ge a+b+c-2\)

mặt khác \(\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ca\)

\(\Rightarrow a+b+c\ge3\)

\(\Rightarrow P\ge3-2=1\)

Dấu = xảy ra a=b=c=1