1. cho \(0< a\le b\le c\) . Cmr: \(\frac{2a^2}{b^2+c^2}+\frac{2b^2}{c^2+a^2}+\frac{2c^2}{a^2+b^2}\le\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
2. cho \(a,b,c\ge0\). cmr: \(a^2+b^2+c^2+3\sqrt[3]{\left(abc\right)^2}\ge2\left(ab+bc+ca\right)\)
3. \(a,b,c>0.\) Cmr: \(\sqrt{\left(a^2b+b^2c+c^2a\right)\left(ab^2+bc^2+ca^2\right)}\ge abc+\sqrt[3]{\left(a^3+abc\right)\left(b^3+abc\right)\left(c^3+abc\right)}\)
4. \(a,b,c>0\). Tìm Min \(P=\left(\frac{a}{a+b}\right)^4+\left(\frac{b}{b+c}\right)^4+\left(\frac{c}{c+a}\right)^4\)
Cho a ,b ,c là các số thực dương thỏa mãn a+b+c+\(\sqrt[]{2abc}\)=2 CMR
\(\sqrt{a\left(2-b\right)\left(2-c\right)}+\sqrt{b\left(2-a\right)\left(2-c\right)}+\sqrt{c\left(2-a\right)\left(2-b\right)}=\sqrt{8}+\sqrt{abc}\)
1. a) cho \(1\le a,b,c\le2\). Tìm max \(P=\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\)
b) \(\left\{{}\begin{matrix}a,b,c\ge0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\sqrt{\frac{3a^2+1}{3b^2+1}}+\sqrt{\frac{3b^2+1}{3c^2+1}}+\sqrt{\frac{3c^2+1}{3a^2+1}}\le\frac{7}{2}\)
2.a) \(a,b\ge0;c\ge1;a+b+c=2\). cmr: \(\left(6-a^2-b^2-c^2\right)\left(2-abc\right)\le8\)
b) \(\left\{{}\begin{matrix}a+b\le2\\a^2+b^2+ab=3\end{matrix}\right.\). Tìm max,min \(P=a^2+b^2-ab\)
Cho a,b,c > 0. Cmr: \(\frac{a\left(b+c\right)}{a^2+\left(b+c\right)^2}+\frac{b\left(c+a\right)}{b^2+\left(c+a\right)^2}+\frac{c\left(a+b\right)}{c^2+\left(a+b\right)^2}\le\frac{6}{5}\)
CMR: Với các số thực dương a;b;c thì\(\dfrac{a^3+2abc+b^3}{c^2+ab}+\dfrac{a^3+2abc+c^3}{b^2+ac}+\dfrac{b^3+2abc+c^3}{a^2+bc}\ge2\left(a+b+c\right)\)
Cho a, b, c > 0 thoã mãn: ab + bc + ca = 3. CMR: \(\dfrac{1}{1+a^2\left(b+c\right)}+\dfrac{1}{1+b^2\left(c+a\right)}+\dfrac{1}{1+c^2\left(a+b\right)}\le\dfrac{3}{abc}\)
Cho a, b, c >0. CMR: \(\dfrac{a+b+c}{3}\) - \(\sqrt[3]{abc}\) ≤ \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2}{3}\)
a) cmr (ax+by+cz)\(^2\)≤\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
b) cho a,b,c >0 tm \(a^2+b^2+c^2=1\)
cmr :\(\frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}\le\frac{a}{2\left(a+b+c\right)}\)
cho a,b,c là các số thực dương thay đổi bất kì
cm:
\(\dfrac{\left(2a+b+c\right)^2}{2a^2+\left(b+c\right)^2}+\dfrac{\left(a+2b+c\right)^2}{2b^2+\left(c+a\right)^2}+\dfrac{\left(a+b+2c\right)^2}{2c^2+\left(a+b\right)^2}\le8\)