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Cho a,b,c>0 t/m: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2011}\)
CMR:\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\sqrt{\frac{2011}{2}}\)
cho ba số duong a,b,c tỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2=\sqrt{2011}}\)chứng minh rằng \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\sqrt{\frac{2011}{2}}\)
với a,b,c dương thỏa
\(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2015}\\CMR:\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{1}{2}\sqrt{\frac{2015}{2}}\)
Cho 3 số dương a;b;c thỏa mãn: \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=1\)
CMR: \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2\sqrt{2}}\)
sqrt{ab}+sqrt{bc}+sqrt{ac}ge3sqrt[6]{abc}3Ta có frac{a}{a+2}+frac{b}{b+2}+frac{c}{c+2}gefrac{left(sqrt{a}+sqrt{b}+sqrt{c}right)^2}{a+b+c+6}frac{a+b+c+2left(sqrt{ab}+sqrt{bc}+sqrt{ac}right)}{a+b+c+6}ge1 frac{a}{a+2}+frac{b}{b+2}+frac{c}{c+2}ge1 left(frac{1}{2}-frac{1}{a+2}right)+left(frac{1}{2}-frac{1}{b+1}right)+left(frac{1}{2}-frac{1}{c+1}right)gefrac{1}{2} frac{1}{a+2}+frac{1}{b+2}+frac{1}{c+2}le1(ĐPCM)
Đọc tiếp
\(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\ge3\sqrt[6]{abc}=3\)
Ta có \(\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+6}=\frac{a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{a+b+c+6}\ge1\)
=> \(\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge1\)
=> \(\left(\frac{1}{2}-\frac{1}{a+2}\right)+\left(\frac{1}{2}-\frac{1}{b+1}\right)+\left(\frac{1}{2}-\frac{1}{c+1}\right)\ge\frac{1}{2}\)
=> \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\le1\)(ĐPCM)
Cho a,b,c>0. Cmr:
\(\frac{a}{\sqrt{ab+b^2}}+\frac{b}{\sqrt{bc+b^2}}+\frac{c}{\sqrt{ac+c^2}}\ge\frac{3\sqrt{2}}{2}\)
Cho a, b, c là các số dương thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\sqrt{2019}\)
CMR : \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\sqrt{\frac{2019}{8}}\)
Cho a b c dương thỏa mãn a+b+c=3 CMR
\(\frac{a}{\sqrt{b+1}}+\frac{b}{\sqrt{c+1}}+\frac{c}{\sqrt{a+1}}\ge\frac{3\sqrt{2}}{2}\)
CMR
\(\sqrt{a^2+\frac{1}{b}}+\sqrt{b^2+\frac{1}{c}}+\sqrt{c^2+\frac{1}{a}}\ge\frac{\sqrt{97}}{2}\)