Cho a,b,c >0 tm a+b+c=1.Tìm max \(S=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)
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Ta có \(4\sqrt[3]{a+7}\le\frac{a+7+8+8}{3}=\frac{a+23}{3}\)
\(4\sqrt[3]{b+7}\le\frac{b+23}{3}\)
Từ đó ta có
\(4P=4\sqrt[3]{a+7}+4\sqrt[3]{b+7}+4\sqrt[3]{b+7}\)
\(\le\frac{a+b+b+23×3}{3}=\frac{a+2b+23×3}{3}\le24\)
\(\Rightarrow P\le6\)
Đạt được khi a = b = 1
Ta có:
\(\sqrt[3]{a+b}=\sqrt[3]{\frac{9}{4}}.\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\frac{\left(a+b\right)+\frac{2}{3}+\frac{2}{3}}{3}\)
Tương tự:
\(\sqrt[3]{b+c}\le\frac{\left(b+c\right)+\frac{2}{3}+\frac{2}{3}}{3}\)
\(\sqrt[3]{c+a}\le\frac{\left(c+a\right)+\frac{2}{3}+\frac{2}{3}}{3}\)
\(\Rightarrow\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{c+a}\le\sqrt[3]{\frac{9}{4}}.\frac{2\left(a+b+c\right)+4}{3}\)
\(=\sqrt[3]{\frac{9}{4}}.\frac{6}{3}=\sqrt[3]{18}\)
(Dấu "="\(\Leftrightarrow\hept{\begin{cases}a+b=\frac{2}{3}\\b+c=\frac{2}{3}\\c+a=\frac{2}{3}\end{cases}}\)\(\Leftrightarrow a=b=c=\frac{1}{3}\))
Em làm sai tại đây nhé:
\(\sqrt[3]{a+b}=\sqrt[3]{\frac{9}{4}}.\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\sqrt[3]{\frac{9}{4}}.\frac{1}{3}.\left(a+b+\frac{2}{3}+\frac{2}{3}\right)\)
\(1,yz\sqrt{x-1}=yz\sqrt{\left(x-1\right)\cdot1}\le yz\cdot\dfrac{x-1+1}{2}=\dfrac{xyz}{2}\)
\(zx\sqrt{y-2}=\dfrac{zx\cdot2\sqrt{2\left(y-2\right)}}{2\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\\ xy\sqrt{z-3}=\dfrac{xy\cdot2\sqrt{3\left(z-3\right)}}{2\sqrt{3}}\le\dfrac{xyz}{2\sqrt{3}}\)
\(\Leftrightarrow M\le\dfrac{\dfrac{xyz}{2}+\dfrac{xyz}{2\sqrt{2}}+\dfrac{xyz}{2\sqrt{3}}}{xyz}=\dfrac{xyz\left(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\right)}{xyz}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=2\\z-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)
\(2,N^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\\ \Leftrightarrow N^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right)\\ \Leftrightarrow N^2\le6\left(a+b+c\right)=6\sqrt{2}\\ \Leftrightarrow N\le\sqrt{6\sqrt{2}}\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{\sqrt{2}}{3}\)
\(S=\sqrt{a+b+c}+\sqrt{b+c+d}+\sqrt{c+d+a}+\sqrt{d+a+b}\)
\(\le\frac{a+b+c}{\sqrt{3}}+\frac{\sqrt{3}}{4}+\frac{b+c+d}{\sqrt{3}}+\frac{\sqrt{3}}{4}+\frac{c+d+a}{\sqrt{3}}+\frac{\sqrt{3}}{4}+\frac{d+a+b}{\sqrt{3}}+\frac{\sqrt{3}}{4}\)
\(=\sqrt{3}+\frac{3}{\sqrt{3}}\left(a+b+c+d\right)=2\sqrt{3}\)
\(\sqrt{a+b}=\dfrac{\sqrt{2}}{\sqrt{3}}.\sqrt{a+b}.\dfrac{\sqrt{3}}{2}\le\dfrac{\dfrac{2}{3}+a+b}{2}.\dfrac{\sqrt{3}}{\sqrt{2}}\)
\(\text{Tương tự :}\sqrt{b+c}\le\dfrac{\sqrt{3}}{\sqrt{2}}\dfrac{\dfrac{2}{3}+b+c}{2};\sqrt{c+a}\le\dfrac{\sqrt{3}}{\sqrt{2}}\dfrac{\dfrac{2}{3}+c+a}{2}\)
\(\text{Khi đó :}S\le\dfrac{\sqrt{3}}{\sqrt{2}}.\dfrac{2+2\left(a+b+c\right)}{2}=\sqrt{6}\)
\(\text{Vậy maxS=}\sqrt{6}\text{ khi }a=b=c=\dfrac{1}{3}\)