Question

a,Cho a,b,c duong va \(a^2+b^2+c^2\)=3. Tim Min cua P= \(\frac{a^3}{\sqrt{b^2+3}}+\frac{b^3}{\sqrt{c^2+3}}+\frac{c^3}{\sqrt{a^2+3}}\)

b,Cho x,y,z>0 va x+y+z=6. C/m \(8^x+8^y+8^z\ge4^{x+1}+4^{y+1}+4^{z+1}\)

Answer 1

a/

-Cauchy-Schwar 

\(P=\sum\frac{a^4}{a\sqrt{b^2+3}}\ge\frac{\left(\sum a^2\right)^2}{\sum a\sqrt{b^2+3}}\)

Côsi: \(\sum a\sqrt{b^2+3}=\frac{1}{2}\sum2a.\sqrt{b^2+3}\le\frac{1}{2}.\sum\frac{\left(2a\right)^2+b^2+3}{2}=\frac{1}{4}.\left[5\left(a^2+b^2+c^2\right)+3.3\right]=6\)

\(\Rightarrow P\ge\frac{3^2}{6}=\frac{3}{2}\)

Đẳng thức xảy ra khi a = b = c = 1.

b/

Côsi: \(8^x+8^x+64\ge3\sqrt[3]{8^x.8^x.64}=12.4^x\Rightarrow8^x\ge6.4^x-32\)

\(\Rightarrow8^x+8^y+8^z\ge6\left(4^x+4^y+4^z\right)-96\)

\(4^x+4^y+4^z\ge3\sqrt[3]{4^{x+y+z}}=3\sqrt[3]{4^6}=48\)

\(\Rightarrow-2\left(4^x+4^y+4^z\right)\le-96\)

\(\Rightarrow8^x+8^y+8^z\ge6\left(4^x+4^y+4^z\right)-2\left(4^x+4^y+4^z\right)=4^{x+1}+4^{y+1}+4^{z+1}\)