nhờ mn giúp mk bài này vs ạ

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cảm ơn mn nhiều

Đặt \(\left(\sqrt[3]{x};\sqrt[3]{y};\sqrt[3]{z}\right)=\left(a;b;c\right)\) \(\Rightarrow a^6+b^6+c^6=3\)

\(a^6+a^6+a^6+a^6+a^6+1\ge6a^5\)

Tương tự: \(5b^6+1\ge6b^5\) ; \(5c^6+1\ge6c^5\)

Cộng vế với vế: \(18=5\left(a^6+b^6+c^6\right)+3\ge6\left(a^5+b^5+c^5\right)\)

\(\Rightarrow3\ge a^5+b^6+b^5\)

BĐT cần chứng minh: \(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}\ge a^3b^3+b^3c^3+c^3a^3\) 

Ta có:

\(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}\ge\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge a+b+c\) (1)

Mà \(3\left(a+b+c\right)\ge\left(a^5+b^5+c^5\right)\left(a+b+c\right)\ge\left(a^3+b^3+c^3\right)^2\ge3\left(a^3b^3+b^3c^3+c^3a^3\right)\)

\(\Rightarrow a+b+c\ge a^3b^3+b^3c^3+c^3a^3\) (2)

Từ (1);(2) \(\Rightarrow\) đpcm

Chắc dùng Mincowski

Làm chi mà khó hiểu thế. Làm lại bài của Thắng Nguyễn cho dễ hiểu. 

\(P=\left(\frac{1}{x}+\frac{2}{y}+\frac{5}{z}\right)\sqrt{xy+yz+zx}\)

\(\Leftrightarrow P^2=\left(\frac{1}{x}+\frac{2}{y}+\frac{5}{z}\right)^2.\left(xy+yz+zx\right)\)

Đặt \(\hept{\begin{cases}x=\frac{a}{3}\\y=\frac{b}{2}\\z=c\end{cases}}\)thì ta có

\(P^2=\left(\frac{3}{a}+\frac{4}{b}+\frac{5}{c}\right)^2.\left(\frac{ab}{6}+\frac{bc}{2}+\frac{ca}{3}\right)\)

\(=\frac{1}{12}\left(\frac{3}{a}+\frac{4}{b}+\frac{5}{c}\right)^2.\left(2ab+6bc+4ca\right)\)

Ta có: \(\frac{3}{a}+\frac{4}{b}+\frac{5}{c}=\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}\ge12.\sqrt[12]{\frac{1}{a^3.b^4.c^5}}\)

\(\Rightarrow\left(\frac{3}{a}+\frac{4}{b}+\frac{5}{c}\right)^2\ge12^2.\sqrt[12]{\frac{1}{a^6.b^8.c^{10}}}\)

Ta lại có: \(2ab+6bc+4ca\ge12.\sqrt[12]{\left(ab\right)^2.\left(bc\right)^6.\left(ca\right)^4}=12.\sqrt[12]{a^6.b^8.c^{10}}\)(tách y hệt cái trên)

Từ đây ta có: \(P^2\ge\frac{1}{12}.12^2.\sqrt[12]{\frac{1}{a^6.b^8.c^{10}}}.12\sqrt[12]{a^6.b^8.c^{10}}=12^2\)

\(\Rightarrow P\ge12\)

Dấu = xảy ra khi a = b = c hay z = 2y = 3x

đề? \(\left(\frac{1}{x}+\frac{2}{y}+\frac{5}{z}\right)\sqrt{xy+yz+xz}\)

Đặt \(\hept{\begin{cases}a=\frac{x}{3}\\b=\frac{y}{2}\\c=z\end{cases}}\). Do đó, áp dụng BĐT AM-GM ta có:

\(\left(\frac{1}{a}+\frac{2}{b}+\frac{5}{c}\right)^2=\left(\frac{1}{a}+\frac{2}{b}+\frac{5}{c}\right)^2\left(ab+ac+bc\right)\)

\(=\frac{1}{12}\left(\frac{3}{x}+\frac{4}{y}+\frac{5}{z}\right)^2\left(2xy+4xz+6yz\right)\)

\(\ge\frac{1}{12}\cdot12^3\sqrt[12]{x^{-6}y^{-8}z^{-10}x^2y^2x^4z^4y^6z^6}=144\)

Vì vậy \(\frac{1}{a}+\frac{2}{b}+\frac{5}{c}\ge12\) 

Dấu "=" xảy ra khi \(x=y=z=1\) thì P đạt GTNN là 12

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)