Cho a,b,c>0 và a=max{a,b,c}.Tìm min của :
\(S=\dfrac{a}{b}+2\sqrt{1+\dfrac{b}{c}}+3\sqrt[3]{1+\dfrac{c}{a}}\)
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19 tháng 4 2023
\(\dfrac{1}{\sqrt{a^2-ab+b^2}}< =\dfrac{1}{\sqrt{2ab-ab}}=\dfrac{1}{\sqrt{ab}}\)
\(\sqrt{\dfrac{1}{b^2-bc+c^2}}< =\dfrac{1}{\sqrt{bc}};\sqrt{\dfrac{1}{c^2-ac+c^2}}< =\dfrac{1}{\sqrt{ac}}\)
=>P<=1/a+1/b+1/c=3
Dấu = xảy ra khi a=b=c=1
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(\left(a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}\right)\left(1+3+5\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow3\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}\ge a+b+c\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{1}{a}+\dfrac{3^2}{b}+\dfrac{5^2}{c}}\)
\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{\left(1+3+5\right)^2}{a+b+c}}=\dfrac{2}{3}\left(a+b+c\right)+\dfrac{27}{\sqrt{a+b+c}}\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(a+b+c\right)+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{1}{6}\left(a+b+c\right)\)
\(\Rightarrow P\ge3\sqrt[3]{\dfrac{27^2\left(a+b+c\right)}{2^3\left(a+b+c\right)}}+\dfrac{1}{6}.9=15\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;3;5\right)\)
29 tháng 11 2021
\(1,yz\sqrt{x-1}=yz\sqrt{\left(x-1\right)\cdot1}\le yz\cdot\dfrac{x-1+1}{2}=\dfrac{xyz}{2}\)
\(zx\sqrt{y-2}=\dfrac{zx\cdot2\sqrt{2\left(y-2\right)}}{2\sqrt{2}}\le\dfrac{xyz}{2\sqrt{2}}\\ xy\sqrt{z-3}=\dfrac{xy\cdot2\sqrt{3\left(z-3\right)}}{2\sqrt{3}}\le\dfrac{xyz}{2\sqrt{3}}\)
\(\Leftrightarrow M\le\dfrac{\dfrac{xyz}{2}+\dfrac{xyz}{2\sqrt{2}}+\dfrac{xyz}{2\sqrt{3}}}{xyz}=\dfrac{xyz\left(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\right)}{xyz}=\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=2\\z-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\\z=6\end{matrix}\right.\)
29 tháng 11 2021
\(2,N^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\\ \Leftrightarrow N^2\le\left(a+b+b+c+c+a\right)\left(1^2+1^2+1^2\right)\\ \Leftrightarrow N^2\le6\left(a+b+c\right)=6\sqrt{2}\\ \Leftrightarrow N\le\sqrt{6\sqrt{2}}\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{\sqrt{2}}{3}\)
17 tháng 11 2021
\(S=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\sqrt{a^2+\dfrac{1}{b^2}}=\dfrac{1}{\sqrt{17}}\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(1+4^2\right)}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\left(1\right)\)\(\left(bunhia\right)\)
\(tương-tự\Rightarrow\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\left(2\right)\)
\(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}+b+\dfrac{4}{c}+c+\dfrac{4}{a}\right)\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[16a+\dfrac{4}{a}+16b+\dfrac{4}{b}+16c+\dfrac{4}{c}-15\left(a+b+c\right)\right]\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[2\sqrt{16a.\dfrac{4}{a}}+2\sqrt{16b.\dfrac{4}{b}}+2\sqrt{16c.\dfrac{4}{c}}-15.\dfrac{3}{2}\right]\left(am-gm\right)\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(16+16+16-\dfrac{45}{2}\right)=\dfrac{3\sqrt{17}}{2}\)
\(\Rightarrow MinS=\dfrac{3\sqrt{17}}{2}\Leftrightarrow a=b=c=\dfrac{1}{2}\)
mong mn giúp mk vs