Ta luôn có \(4\left(x^3+y^3\right)\ge\left(x+y\right)^3\)(*)

Thật vậy: (*)\(\Leftrightarrow3\left(x-y\right)^2\left(x+y\right)\ge0\)*Đúng với mọi x, y thực dương*

\(\Rightarrow\sqrt[3]{4\left(x^3+y^3\right)}\ge x+y\)

Tương tự, ta có: \(\sqrt[3]{4\left(y^3+z^3\right)}\ge y+z,\sqrt[3]{4\left(z^3+x^3\right)}\ge z+x\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(\sqrt[3]{4\left(x^3+y^3\right)}+\sqrt[3]{4\left(y^3+z^3\right)}+\sqrt[3]{4\left(z^3+x^3\right)}\ge2\left(x+y+z\right)\)

Ta cần chứng minh \(\left(x+y+z\right)+\left(\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\right)\ge6\)

Thật vậy, ta có: \(\left(x+y+z\right)+\left(\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\right)\ge3\sqrt[3]{xyz}+\frac{3}{\sqrt[3]{xyz}}\ge3.2=6\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi x = y = z 

Ta có :a² +b²+c²  =5  (a+b+c)-2ab <=> (a+b)²+c²=5(a+b+c)

Áp dụng bđt bunhiacopxki có:

(a+b)² +c² > 1/2 (a+b+c)²

=> 1/2 (a+b+c)² < 5(a+b+c) => 0< a+b+c <10

Áp dụng bđt Cauchy ta có:

\(\sqrt{\frac{3}{\sqrt{a+10}}}=\frac{1}{\sqrt{\frac{a+10}{3}}}\):\(\sqrt{\frac{a+10}{3}}\)

=\(\frac{1}{2}\sqrt{\frac{a+10}{3}\cdot}4\)<  1/4 (\(\frac{a+10}{3}+4\))

= a+22/12 => \(\frac{\sqrt{3}}{\sqrt{a+10}}\)>  \(\frac{12}{â+22}\)

\(\sqrt[3]{b+c}=\frac{1}{4}\sqrt[3]{\left(b+c\right)8.8}\)< \(\frac{1}{4}\cdot\frac{b+c+8+8}{3}\)

=\(\frac{b+c+16}{12}\)=> \(\frac{1}{\sqrt[3]{b+c}}\)> \(\frac{12}{b+c+16}\)

=> P > a= b+c+48 . 12 (\(\frac{1}{a+22}+\frac{1}{b+c+16}\))

Áp dụng bđt Cauchy - Schwarz ta được :

\(\frac{1}{a+22}+\frac{1}{b+c+16}\)> \(\frac{4}{a+b+c}+38\)=> P>  a+b+c+\(\frac{2304}{a+b+c+38}\)

Đặt t= a+b+c => t\(\in\)(0;10) => P> t+\(\frac{2304}{t+38}\)

Xét hàm f(t) = t+\(\frac{2304}{t+38}\)trên ( 0;10)

Ta có : f(t) =1-\(\frac{2304}{\left(t+38\right)^2}=\frac{\left(t-10\right)\left(t+86\right)}{\left(t+38\right)^2}\)

=> f(t)< 0 \(\forall\)t \(\in\)( 0:10)

=> f(t) nghịch biến trên (0;10) => f(t) >  f(10) \(\forall\)t \(\in\)(0;10); f(10) =58 => P>58

Daaus "=" xảy ra khi \(\hept{\begin{cases}a+b+c=10\\a+b=c\\\frac{a+10}{3}\end{cases}}\)hoặc b+c=8 \(\hept{\begin{cases}a=2\\b=3\\c=5\end{cases}}\)

Vậy min P =58 khi a=2 , b=3 , c=5 

HỌC TỐT 

\(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\Leftrightarrow\frac{y+x}{xy}=\frac{1}{z}\Leftrightarrow xz+yz=xy\Leftrightarrow2xz+2yz-2xy=0\)

Tac có : \(\sqrt{x^2+y^2+z^2}=\sqrt{x^2+y^2+z^2+2xz+2yz-2xy}\)

\(=\sqrt{\left(x+y-z\right)^2}=\left|x+y-z\right|\)

\(\Rightarrow\sqrt{x^2+y^2+z^2}\)là số hữu tỉ ( đpcm )

Ta có: \(4\left(x^2-x+1\right)\le16\sqrt{x^2yz}-3x\left(y+z\right)^2\le16x.\frac{y+z}{2}-3x\left(y+z\right)^2=8x\left(y+z\right)-3x\left(y+z\right)^2\)\(\Leftrightarrow4\left(x+\frac{1}{x}\right)-4\le8\left(y+z\right)-3\left(y+z\right)^2\)

Mà \(x+\frac{1}{x}\ge2\left(Cauchy\right)\)nên \(8\left(y+z\right)-3\left(y+z\right)^2\ge4\Leftrightarrow\frac{2}{3}\le y+z\le2\)

Có\(2\ge y+z\ge2\sqrt{yz}\Rightarrow yz\le1\)

\(P=\frac{y^2+3xy\left(x+1\right)}{x^2.yz}+\frac{16}{\left(y+1\right)^3}-10\sqrt{\frac{3y}{x^3+1+1}}\)\(\ge\frac{y^2+3xy\left(x+1\right)}{x^2}+\frac{16}{\left(y+1\right)^3}-10\sqrt{\frac{y}{x}}=\left(\frac{y}{x}\right)^2+3\left(\frac{y}{x}\right)\)\(-10\sqrt{\frac{y}{x}}+3y+\frac{16}{\left(y+1\right)^3}=\left(\frac{y}{x}\right)^2+3\left(\frac{y}{x}\right)-10\sqrt{\frac{y}{x}}+6+\left(y+1\right)+\left(y+1\right)+\left(y+1\right)+\frac{16}{\left(y+1\right)^3}-9\)\(\ge\left(\sqrt{\frac{y}{x}}-1\right)^2\left(\frac{y}{x}+2\sqrt{\frac{y}{x}}+6\right)+4\sqrt[4]{\frac{16\left(y+1\right)^3}{\left(y+1\right)^3}}-9\ge-1\)

Vậy MinP = -1 khi x = y = z = 1