\(\sqrt[3]{abc}+\sqrt[3]{xyz}\le\sqrt[3]{\left(a+x\right)\left(b+y\right)\left(c+z\right)}\)
\(\Leftrightarrow\sqrt[3]{\dfrac{abc}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}+\sqrt[3]{\dfrac{xyz}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}\le1\)
Áp dụng BĐT Cô-si cho 3 số dương, ta có:
\(\Leftrightarrow\sqrt[3]{\dfrac{abc}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}+\sqrt[3]{\dfrac{xyz}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}\le\dfrac{1}{3}\left(\dfrac{a}{a+x}+\dfrac{b}{b+y}+\dfrac{c}{c+z}+\dfrac{x}{a+x}+\dfrac{y}{b+y}+\dfrac{z}{c+z}\right)=1\)
\(\sqrt[3]{abc}\le\dfrac{a+b+c}{3}\)
\(\sqrt[3]{xyz}\le\dfrac{x+y+z}{3}\)
\(\Rightarrow\sqrt[3]{abc}+\sqrt[3]{xyz}\le\dfrac{\left(a+x\right)+\left(b+y\right)+\left(c+z\right)}{3}\)
Áp dụng BĐT Cô-si cho 3 số dương (a+x); (b+y); (c+z) , ta có:
\(\sqrt[3]{\left(a+x\right)\left(b+y\right)\left(c+z\right)}\le\dfrac{\left(a+x\right)}{ }\)
câu này cô bảo là còn cách ngắn hơn xin hãy giúp cho ạ?
\(\sqrt[3]{abc}+\sqrt[3]{xyz}\ge\sqrt[3]{\left(a+x\right)\left(b+y\right)\left(c+z\right)}\)
Áp dụng bđt cosi có dạng :
a+b=\(2\sqrt{ab}\)
\(\sqrt[3]{\dfrac{abc}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}+\sqrt[3]{\dfrac{xyz}{\left(a+x\right)\left(b+y\right)\left(c+z\right)}}\ge1\)
\(\dfrac{\dfrac{a}{a+x}+\dfrac{b}{b+y}+\dfrac{c}{c+z}+\dfrac{x}{a+x}+\dfrac{y}{b+y}+\dfrac{z}{c+z}}{3}\)\(\dfrac{\left(\dfrac{a}{a+x}+\dfrac{x}{a+x}\right)+\left(\dfrac{b}{b+y}+\dfrac{y}{b+y}\right)+\left(\dfrac{c}{c+z}+\dfrac{z}{c+z}\right)}{3}=\dfrac{\dfrac{a+x}{a+x}+\dfrac{b+y}{b+y}+\dfrac{c+z}{c+z}}{3}=\dfrac{1+1+1}{3}=1.\Rightarrow\sqrt[3]{abc}+\sqrt[3]{xyz}\ge\sqrt[3]{\left(a+x\right)\left(b+y\right)\left(c+z\right)}\)
