cho 3 so duong a,b,c thoa man a+b+c=1/abc chung minh rang can ((1+b^2c^2)(1+a^2c^2)/c^2+a^2b^2c^2)=a+b
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cho a,b,c la cac so thuc duong. chung minh rang 2a/(b+c)+2b/(c+a)+2c/(a+b)>=((a-b)^2+(b-c)^2+(c-a)^2)/(a+b+c)^2
cho a , b, c la cac so thuc duong thoa man he thuc a+b+c=6abc
Chung minh rang \(\dfrac{bc}{a^3\left(c+2b\right)}+\dfrac{ac}{b^3\left(a+2c\right)}+\dfrac{ab}{c^3\left(b+2a\right)}\ge2\)
Cho a,b.c la cac so duong va abc = 1
Chung minh rang \(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\)
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\(\ge0\)thoa man abc=1.Chung minh rang
\(\frac{1}{2a^3+3a+2}+\frac{1}{2b^3+3b+2}+\frac{1}{2c^3+3c+2}\)\(\ge\frac{3}{7}\)
Cho a, b,c : abc = 1. Chứng minh:
\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}+\dfrac{b^2c^2}{2b^2+c^2+3b^2c^2}+\dfrac{c^2a^2}{2c^2+a^2+3a^2c^2}\le\dfrac{1}{2}\)
\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)
Tương tự và cộng lại;
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Đúng 1
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cho abc=36,1/a+1/b+1/c=o.
Tính Q=a^2(b^2+c^2)-b^2c^2/a^2b^2c^2*b^2(c^2+a^2)-c^2a^2/a^2b^2c^2*c^2(a^2+b^2)-a^2b^2/a^2b^2c^2
Cho \(A=\frac{a^2\left(b^2+c^2\right)-b^2c^2}{a^2b^2c^2}\) ; \(B=\frac{b^2\left(a^2+c^2\right)-a^2c^2}{a^2b^2c^2}\) ; \(C=\frac{c^2\left(a^2+b^2\right)-a^2b^2}{a^2b^2c^2}\)
Và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). Tính ABC
Cho \(A=\frac{a^2\left(b^2+c^2\right)-b^2c^2}{a^2b^2c^2}\) ; \(B=\frac{b^2\left(a^2+c^2\right)-a^2c^2}{a^2b^2c^2}\) ; \(c=\frac{c^2\left(a^2+b^2\right)-a^2b^2}{a^2b^2c^2}\)
Và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). Tính ABC