Cho a,b>0,a+b=1.Tim GTNN cua A=\(\frac{3}{a^2+b^2}+\frac{2}{ab}\)
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cho a,b>0(a+b<=1) tim GTNN cua J=\(\frac{1}{a^2+b^2}+\frac{1}{ab}\)
\(J=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}+\frac{1}{\frac{2\left(a+b\right)^2}{4}}=\frac{6}{\left(a+b\right)^2}\ge6\)
\(\Rightarrow J_{min}=6\) khi \(a=b=\frac{1}{2}\)
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Cho 2 so thuc duong a,b thoa man a+b<=1.Tim GTNN cua
\(A=\frac{1}{a^3+b^3}+\frac{1}{a^2b}+\frac{1}{ab^2}\)
Cho a,b,c > 0 . Tim GTNN cua P =\(\frac{a^2}{a-1}+\frac{2b^2}{b-1}+\frac{3c^2}{c-1}\)
Cho a,b,c > 0 . Tim GTNN cua P = \(\frac{a^2}{a-1}+\frac{2b^2}{b-1}+\frac{3c^2}{c-1}\)
Cho a,b>0 va a+b nho hon hoac bang 1. Tim GTNN \(S=\frac{1}{a^3+b^3}+\frac{1}{a^2b}+\frac{1}{ab^2}\)
cho a,b>0.Tim GTNN cua \(P=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}.\)
áp dụng bdt cô-si ta có P\(\ge\)2
dấu = xảy ra khi (a+b)2=ab
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\(\text{Giải}\)
\(P=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)
Ấp dụng BĐT Cô-si ta có:
\(a+b\ge2\sqrt{ab}\)
\(P=\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}+\frac{a+b}{\sqrt{ab}}.\frac{3}{4}\)
\(\text{ÁP DỤNG BĐT Cô-si Ta đc:}\)\(\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\ge2\sqrt{\frac{\left(a+b\right)\left(\sqrt{ab}\right)}{4\sqrt{ab}\left(a+b\right)}}=1\)
Theo BĐT Cô si ta đc:\(\frac{3}{4}.\frac{a+b}{\sqrt{ab}}\ge\frac{3}{4}.2=\frac{3}{2}\)
\(\Rightarrow P_{min}=\frac{3}{2}.\text{Dấu "=" xảy ra khi: a=b}\)
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Xem thêm câu trả lời
Cho a>1 b>1 Tim GTNN cua \(A=\frac{a^2}{a-1}+\frac{b^2}{b-1}\)
cho a,b,c>0 va a+b+c=1
Tim GTNN cua \(P=\sqrt{\frac{ab}{c+ab}}+\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ac}{b+ac}}....\)
cho 3 so duong a;b;c thoa man a+b+c=1.tim GTNN cua:
\(p=\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)