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Cho a,b,c>1. Tìm GTNN của:
\(\frac{a}{\sqrt{b}-1}\)+\(\frac{b}{\sqrt{c}-1}\)+\(\frac{c}{\sqrt{a}-1}\)
S = a+b+c + (1/a + 1/b + 1/c)
>= (a+b+c) + 9/a+b+c
= [ (a+b+c) + 9/4.(a+b+c) ] + 27/4.(a+b+c)
>= \(2\sqrt{\left(a+b+c\right).\frac{9}{4.\left(a+b+c\right)}}\) + 27/(4.3/2)
= 3 + 9/2
= 15/2
Dấu "=" xảy ra <=> a=b=c=1/2
Vậy ......
Tk mk nha
Áp dụng bđt Caauchy ta có :
\(a+b+c\ge3\sqrt[3]{abc}\Leftrightarrow3\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)
\(\Rightarrow P=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{3}=3\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{3}=3\)
minP=3 khi a=b=c=1
\(P=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2\left(b+c\right)}+\frac{ac}{b^2\left(a+c\right)}+\frac{ab}{c^2\left(a+b\right)}\left(abc=1\right)\)
\(=\frac{1}{a^2\left(\frac{1}{c}+\frac{1}{b}\right)}+\frac{1}{b^2\left(\frac{1}{c}+\frac{1}{a}\right)}+\frac{1}{c^2\left(\frac{1}{b}+\frac{1}{a}\right)}\)
\(=\frac{\frac{1}{a^2}}{\frac{1}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{b}+\frac{1}{a}}\)
Đặt \(\left\{\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\) suy ra \(xyz=1\). Khi đó:
\(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
Áp dụng BĐT AM-GM ta có:
\(\left\{\begin{matrix}\frac{x^2}{y+z}+\frac{y+z}{4}\ge x\\\frac{y^2}{x+z}+\frac{x+z}{4}\ge y\\\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\end{matrix}\right.\).Cộng theo vế ta có:
\(P+\frac{x+y+z}{2}\ge x+y+z\)
\(\Rightarrow P\ge\frac{x+y+z}{2}\ge\frac{3}{2}\left(x+y+z\ge3\sqrt[3]{xyz}=3\right)\)
\(P=\frac{1^2}{a}+\frac{1^2}{b}+\frac{2^2}{c}+\frac{4^2}{d}\ge\frac{\left(1+1+2+4\right)^2}{a+b+c+d}=\frac{8^2}{8}=8\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}a+b+c+d=8\\\frac{1}{a}=\frac{1}{b}=\frac{4}{c}=\frac{16}{d}\end{cases}}\)
\(P=\frac{a+b}{abc}=\frac{1}{c}\left(\frac{a+b}{ab}\right)=\frac{1}{1-\left(a+b\right)}.\left(\frac{1}{a}+\frac{1}{b}\right)\ge\frac{1}{\left(1-2\sqrt{ab}\right)}.\frac{2}{\sqrt{ab}}\)
\(P\ge\frac{4}{\left(1-2\sqrt{ab}\right).2\sqrt{ab}}\ge\frac{4}{\frac{\left(1-2\sqrt{ab}+2\sqrt{ab}\right)^2}{4}}=16\)
\(\Rightarrow P_{min}=16\) khi \(\left\{{}\begin{matrix}a=b=\frac{1}{4}\\c=\frac{1}{2}\end{matrix}\right.\)