cho a,b,c>0. CMR
\(\frac{2ab}{3a+8b+6c}+\frac{3bc}{3b+6c+4}+\frac{3ac}{9c+4a+4b}\le\frac{a+2b+3c}{2}\)
Những câu hỏi liên quan
Cho a,b,c>0 thỏa mãn a+2b+3c=1
CMR: \(\frac{2ab}{a^2+4b^2}+\frac{6bc}{4b^2+9c^2}+\frac{3ac}{9c^2+a^2}+\frac{1}{4}\left(\frac{1}{a}+\frac{1}{2b}+\frac{1}{3c}\right)\ge\frac{15}{4}\)
Cho a,b,c > 0.CMR:
\(\frac{ab}{a^2+3b^2+4ab+5bc+3ac}+\frac{bc}{2a^2+b^2+3c^2+3ab+4bc+5ac}+\frac{ac}{3a^2+2b^2+c^2+5ab+3bc+4ac}\le\frac{1}{6}\)
Cho a, b, c là các số dương . CMR:
\(\frac{a\left(b+2c\right)}{\sqrt{3b^2+6c^2}}+\frac{b\left(c+2a\right)}{\sqrt{3c^2+6a^2}}+\frac{c\left(a+2b\right)}{\sqrt{3a^2+6b^2}}\le a+b+c\)
ta có:
\(\left(b-c\right)^2\ge0\Leftrightarrow b^2+4bc+4c^2\le3b^2+6c^2\Leftrightarrow\left(b+2c\right)^2\le3b^2+6c^2\)
\(\Leftrightarrow\frac{\left(b+2c\right)^2}{3b^2+6c^2}\le1\Leftrightarrow\frac{b+2c}{\sqrt{3b^2+6c^2}}\le1\Leftrightarrow\frac{a\left(b+2c\right)}{\sqrt{3b^2+6c^2}}\le a\)
cmtt =>\(\frac{a\left(b+2c\right)}{\sqrt{3b^2+6c^2}}+\frac{b\left(c+2a\right)}{\sqrt{3c^2+6a^2}}+\frac{c\left(a+2b\right)}{\sqrt{3a^2+6b^2}}\le a+b+c\left(Q.E.D\right)\)
dấu = xảy ra khi a=b=c
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cho a;b;c;x;y;z khác 0 thỏa mãn:
\(\frac{x^2-6xy}{a}=\frac{4y^2-3xz}{2b}=\frac{9z^2-2xy}{3c}\)
CMR:
\(\frac{a^2-6bc}{x}=\frac{4b^2-3ac}{2y}=\frac{9c^2-2ab}{3z}\)
cho a;b;c;x;y;z khác 0 thỏa mãn:
\(\frac{x^2-6xy}{a}=\frac{4y^2-3xz}{2b}=\frac{9z^2-2xy}{3c}\)
CMR:
\(\frac{a^2-6bc}{x}=\frac{4b^2-3ac}{2y}=\frac{9c^2-2ab}{3z}\)
Cho a,b,c thỏa (a+2b)(2b+3c)(3c+a)#0 và
\(\frac{a^2}{a+2b}+\frac{4b^2}{2a+3b}+\frac{9c^2}{3c+a}=\frac{a^2}{2b+3c}+\frac{4b^2}{3c+a}+\frac{9c^2}{a+2b}\)
chứng minh rằng \(\frac{a}{6}=\frac{b}{3}=\frac{c}{2}\).mấy a giải giúp em cái
chi a,,b,c thoa man (a+2b)(2b+3c)(3c+a)khac 0 va
\(\frac{a^2}{a+2b}+\frac{4b^2}{2b+3c}+\frac{9c^2}{3c+a}=\frac{a^2}{2b+3c}+\frac{4b^2}{a+3c}+\frac{9c^2}{a+2b}\)
cmr;\(\frac{a}{6}=\frac{b}{3}=\frac{c}{2}\)
Cho a,b,c là các số thực dương thỏa mãn a + b + c = 2. CMR
\(\frac{a}{4a+3bc}+\frac{b}{4b+3ac}+\frac{c}{4c+3ab}\) ≤ \(\frac{1}{2}\)
\(\Leftrightarrow\frac{4a}{4a+3bc}+\frac{4b}{4b+3ac}+\frac{4c}{4c+3ab}\le2\)
\(\Leftrightarrow\frac{bc}{4a+3bc}+\frac{ac}{4b+3ac}+\frac{ab}{4c+3ab}\ge\frac{1}{3}\)
Thật vậy, ta có:
\(VT=\frac{b^2c^2}{4abc+3b^2c^2}+\frac{a^2c^2}{4abc+3a^2c^2}+\frac{a^2b^2}{4abc+3a^2b^2}\)
\(VT\ge\frac{\left(ab+bc+ca\right)^2}{3\left(a^2b^2+b^2c^2+c^2a^2\right)+12abc}=\frac{a^2b^2+b^2c^2+c^2a^2+2\left(a+b+c\right)abc}{3\left(a^2b^2+b^2c^2+c^2a^2+4abc\right)}\)
\(VT\ge\frac{a^2b^2+b^2c^2+c^2a^2+4abc}{3\left(a^2b^2+b^2c^2+c^2a^2+4abc\right)}=\frac{1}{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)
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cho a,b,c > 0. Cmr: \(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\frac{a+b+c}{6}\)
