Có:
\(\left\{{}\begin{matrix}a^3+b^3\ge ab\left(a+b\right)\\b^3+c^3\ge bc\left(b+c\right)\\c^3+a^3\ge ca\left(c+a\right)\end{matrix}\right.\Rightarrow VT\ge a^2.\frac{b+c}{2}+b^2.\frac{c+a}{2}+c^2.\frac{a+b}{2}\ge a^2\sqrt{bc}+b^2\sqrt{ca}+c^2\sqrt{ab}=VP\)
Dấu bằng xảy ra khi a=b=c
Cho a, b, c. CMR:
\(\sqrt[3]{\left(\frac{2a}{b+c}\right)^2}+\sqrt[3]{\left(\frac{2b}{c+a}\right)^2}+\sqrt[3]{\left(\frac{2c}{a+b}\right)^2}\ge3\)
\(\sqrt[4]{a^3}\)+\(\sqrt[4]{b^3}\)+\(\sqrt[4]{c^3}\)>2\(\sqrt{2}\)
biết a, b, c là các số dương
tính a) \(\sqrt[3]{2-\sqrt{5}}.\left(\sqrt[6]{9+4\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}\right)\)
b) \(\sqrt[4]{17+12\sqrt{2}}-\sqrt{2}\)
c) \(\sqrt[4]{56-24\sqrt{5}}\)
d) \(1+\sqrt[4]{28-16\sqrt{3}}\)
e) \(\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}\)
@Akai Haruma giúp em với chị ơi
Cho a,b,c >0 và a=max{a,b,c} .Tìm gtnn của :
\(S=\dfrac{a}{b}+2\sqrt{1+\dfrac{b}{c}}+3\sqrt[3]{1+\dfrac{c}{a}}\)
Rút gọn:
A=\(\dfrac{\sqrt{a}+2}{\sqrt{a+3}}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
B=\((1-\dfrac{\sqrt{a}}{\sqrt{a}+1})\div(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}+\dfrac{\sqrt{a}+2}{3-\sqrt{a}}+\dfrac{\sqrt{a}+2}{a-5\sqrt{a}+6)}\)
C=\((\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{3\sqrt{a}+1}+\dfrac{8\sqrt{a}}{9a-1})\div(1-\dfrac{3\sqrt{a}-2}{3\sqrt{a}+1)}\)
Giups mình cái mình đag cần gấp
Cho a,b,c∈R.CM bđt \(a^2+b^2+c^2\ge ab+bc+ca\) (1). Áp dụng cm các bđt sau:
a)\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
b)\(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\)
c)\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
d)\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
e)\(\frac{a+b+c}{3}\ge\sqrt{\frac{ab+bc+ca}{3}}vớia,b,c>0\)
f)\(a^4+b^4+c^4\ge abc\) nếu a+b+c=1
cho a,b,c>0 thỏa mãn ab+bc+ca=3.cmr:
\(P=\frac{a^2b+b^2c+c^2a}{3}+\sqrt[3]{9\left(a+b+c\right)}\ge4\)
Rút gọn các biểu thức sau
a. \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
b. \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
c. \(\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
Thực hiện các phép tính
a, \(\sqrt{5+2\sqrt{6}}-\sqrt{2-2\sqrt{6}}\)
b,\(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
c, \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
d, \(\sqrt{24+8\sqrt{5}+}\sqrt{9-4\sqrt{5}}\)
