cho a,b,c >0.Thoa man a+b+c=3.Tim GTNN cua a^2+b^2+c^3
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cho cac so duong a,b,c thoa man : ab+a+b=3
tim GTNN cua bieu thuc C=a^2+b^2
cho 3 so thuc duong a, b, c thoa man 1/a+1/c=2/b. tim GTNN cua (a+b)/(2a-b)+(b+c)(/2c-b)
\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b=\frac{2ac}{a+c}\)
\(P=\frac{a+b}{2a-b}+\frac{b+c}{2c-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{\frac{2ac}{a+c}+c}{2c-\frac{2ac}{a+c}}=\frac{a+3c}{2a}+\frac{3a+c}{2c}=1+\frac{3}{2}\left(\frac{a}{c}+\frac{c}{a}\right)\ge4\)
Dấu "=" xảy ra khi \(a=b=c\)
cho a,b,c la cac so thoa man (a+1)^2+(b+2)^2+(c+3)2<2010.tim GTNN cua bieu thuc A=ab+b(c-1)+c(a-2)
cho a,b,c la cac so thoa man a^2+b^2+c^2=<8 tim GTNN cua xy+yz+2xz
http://olm.vn/hoi-dap/question/595391.html
Bài giải đây bạn nhé! Mà bạn xem lại đề bài , sao lại từ a,b,c lại chuyển qua x,y,z vậy?
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Không có chi! ^.^
Chúc bạn học tốt :))
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cho a,b>0 thoa man a+b+c=6.Tim GTNN cua \(P=\frac{a^3}{\left(a+b\right)\left(b+2c\right)}+\frac{b^3}{\left(b+c\right)\left(c+2a\right)}+\frac{c^3}{\left(c+a\right)\left(a+2b\right)},\)
Bạn cho mình hỏi là chỉ a,b > 0 hay cả a,b,c > 0 vậy
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Cho a, b, c > 0 thoa man a + b + c = 3.
Tim GTNN : \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(VT=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{9}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(=\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}+\dfrac{1}{a^2+b^2+c^2}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ac+ab+bc+ac+a^2+b^2+c^2}+\dfrac{7}{ab+bc+ac}\)
\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\)
Áp dụng bất đẳng thức AM-GM cho 2 số dương:
\(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1^2}{3}=\dfrac{1}{3}\)
Ta có: \(\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
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Cho a, b, c > 0 thoa man a + b + c = 3.
Tim GTNN : \(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\)
Áp dụng BĐT Cauchy-Schwarz ta có
BT\(\ge\)\(\frac{\left(1+1+1\right)^2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}=\frac{9}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}\)
\(=\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}+\frac{7}{ab+bc+ac}\)
\(\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}+\frac{7}{ab+bc+ac}\)\(=1+\frac{7}{ab+bc+ac}\)
Ta lại có ab+bc+ac =< (a+b+c)^2/3 =3
\(\Rightarrow BT\ge1+\frac{7}{3}=\frac{10}{3}\)
Vậy GTNN là \(\frac{10}{3}\)khi a=b=c=1
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cho 3 so duong a;b;c thoa man a+b+c=1.tim GTNN cua:
\(p=\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)
Cho 3 so a,b,c thoa man (a + b + c)2 = 3(a2 + b2 + c2). Tim GTNN P = a2 + (a + 2)(b + c) + 2020
\(\left(a+b+c\right)^2=3a^2+3b^2+3c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)
\(\Rightarrow P=a^2+\left(a+2\right)\left(a+a\right)+2020\)
\(\Rightarrow P=3a^2+4a+2020=3\left(a+\frac{2}{3}\right)^2+\frac{6056}{3}\ge\frac{6056}{3}\)
\(P_{min}=\frac{6056}{3}\) khi \(a=-\frac{2}{3}\)
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