\(a^4+b^4+c^4\ge\frac{\left(a^2+b^2+c^2\right)^2}{3}\ge\frac{\left(\frac{\left(a+b+c\right)^2}{3}\right)^2}{3}=\frac{\left(a+b+c\right)^4}{27}=\frac{1}{27}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
\(\text{Cho a, b, c }\ge0\text{ thỏa mãn }a+b+c=1\)
\(\text{CMR: }a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\ge\frac{\sqrt{3}}{3}\)
\(\text{Cho }a,b,c>0\text{ thỏa mãn }a+b+c=3\)
\(\text{CMR: }\frac{1+b}{1+4a^2}+\frac{1+c}{1+4b^2}+\frac{1+a}{1+4c^2}\ge\frac{6}{5}\)
Cho a,b,c >0 thỏa a + b + c =1.
CMR: \(a^4+b^4+c^4\ge\frac{1}{27}\)
\(\text{Cho }a+b+c=0\text{ thỏa mãn }a^2=\left(2a+2b+2\right)\left(a+c-1\right)\)
\(\text{Cho }\text{a,b}\ge0\text{ thỏa }x^2+4y=8.\text{Tìm Min}:\)
\(\text{A=}x+y+\frac{10}{x+y}\)
Cho \(\left\{{}\begin{matrix}\text{x, y, z > 0}\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{4}\end{matrix}\right.\). Tìm \(\min\limits_P=\dfrac{1}{\alpha\text{a}+\beta b+\gamma c}+\dfrac{1}{\beta\text{a}+\gamma b+\alpha c}+\dfrac{1}{\gamma\text{a}+\alpha b+\beta c} v\text{ới} \alpha; \beta;\text{ \gamma}\in\) \(\mathbb{N}^*\)
cho 4 số thực dương a,b,c,d thỏa mãn a+b+c+d=4.CMR:
\(\frac{1}{ab}+\frac{1}{cd}\ge\frac{a^2+b^2+c^2+d^2}{2}\)
\(\text{Cho }\left(a+2\right)\left(b+2\right)=\frac{25}{4}\)
\(\text{Tìm Min }F=\sqrt{a^4+1}+\sqrt{b^4+1}\)
rút gọn các biểu thức sau
a) \(\frac{4}{\sqrt{10}}\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)\)
b)\(\left(4+\sqrt{\text{15}}\right).\left(\sqrt{10}-\sqrt{6}\right).\sqrt{\text{4}-\sqrt{15}}\)
c)\(\sqrt{\text{4 }\sqrt{\text{6}}\text{ }+8\sqrt{\text{3 }}+4\sqrt{2}+18}\)