Ta có : \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\ge xy\left(x+y\right)\)

\(\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y\right)=xy\left(x+y+z\right)\ge3xy\sqrt[3]{xyz}=3xy\)

\(\frac{\sqrt{1+x^3+y^3}}{xy}\ge\frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)

Tương tự : \(\frac{\sqrt{1+y^3+z^3}}{yz}\ge\frac{\sqrt{3yz}}{yz}=\sqrt{\frac{3}{yz}}\);  \(\frac{\sqrt{1+x^3+z^3}}{xz}\ge\frac{\sqrt{3xz}}{xz}=\sqrt{\frac{3}{xz}}\)

\(\Rightarrow A\ge\sqrt{3}\left(\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{xz}}\right)\ge3\sqrt{3}\sqrt{\frac{1}{\sqrt{x^2y^2z^2}}}=3\sqrt{3}\)

Ta có (x+y)xy=x²+y²-xy

=> \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}\)

<=>\(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^2+\frac{3}{4}\left(\frac{1}{x}-\frac{1}{y}\right)^2\ge\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^2\)

<=> \(0\le\frac{1}{x}+\frac{1}{y}\le4\)

mà \(A=\frac{1}{x^3+y^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^2\le16\)

Vậy Max A =16 khi \(x=y=\frac{1}{2}\)

Đặt \(a=\sqrt{x},b=\sqrt{y},c=\sqrt{z}\left(a,b,c>0\right)\)

Khi đó 

\(P=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)và \(a^2+b^2+c^2\ge3\)

<=>\(P=\frac{a^4}{a^2b}+\frac{b^4}{cb^2}+\frac{c^4}{ac^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2b+cb^2+ac^2}\)(bất đẳng thức cosi schwaz)

Ta có 

\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)=\left(a^3+b^2a\right)+\left(b^3+bc^2\right)+\left(c^3+ca^2\right)+\left(a^2b+b^2c+c^2a\right)\)

                                                        \(\ge3\left(a^2b+b^2c+c^2a\right)\)

=> \(a^2b+b^2c+c^2a\le\frac{1}{3}\left(a+b+c\right)\left(a^2+b^2+c^2\right)\le\frac{\sqrt{3}}{3}\sqrt{\left(a^2+b^2+c^2\right)^3}\)

Khi đó 

\(P\ge\sqrt{3}.\frac{\left(a^2+b^2+c^2\right)^2}{\sqrt{\left(a^2+b^2+c^2\right)^3}}=\sqrt{3\left(a^2+b^2+c^2\right)}\ge3\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1 => x=y=z=1

Em thử ạ!Em không chắc đâu.Hơi quá sức em rồi

Ta có: \(VT=\Sigma\frac{x^3}{z+y+yz+1}=\Sigma\frac{x^3}{z+y+\frac{1}{x}+1}\)

\(=\Sigma\frac{x^4}{xz+xy+1+x}=\frac{x^4}{xy+xz+x+1}+\frac{y^4}{yz+xy+y+1}+\frac{z^4}{zx+yz+z+1}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel,suy ra:

\(VT\ge\frac{\left(x^2+y^2+z^2\right)^2}{\left(x+y+z\right)+2\left(xy+yz+zx\right)+3}\)

\(\ge\frac{\left(\frac{1}{3}\left(x+y+z\right)^2\right)^2}{\left(x+y+z\right)+\frac{2}{3}\left(x+y+z\right)^2+3}\) (áp dụng BĐT \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3};ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))

Đặt \(t=x+y+z\ge3\sqrt{xyz}=3\) Dấu "=" xảy ra khi x = y = z

Ta cần chứng minh: \(\frac{\frac{t^4}{9}}{\frac{2}{3}t^2+t+3}\ge\frac{3}{4}\Leftrightarrow\frac{t^4}{9\left(\frac{2}{3}t^2+t+3\right)}=\frac{t^4}{6t^2+9t+27}\ge\frac{3}{4}\)(\(t\ge3\))

Thật vậy,BĐT tương đương với: \(4t^4\ge18t^2+27t+81\)

\(\Leftrightarrow3t^4-18t^2-27t+t^4-81\ge0\)

Ta có: \(VT\ge3t^4-18t^2-27t+3^4-81\)

\(=3t^4-18t^2-27t\).Cần chứng minh\(3t^4-18t^2-27t\ge0\Leftrightarrow3t^4\ge18t^2+27t\)

Thật vậy,chia hai vế cho \(t\ge3\),ta cần chứng minh \(3t^3\ge18t+27\Leftrightarrow3t^3-18t-27\ge0\)

\(\Leftrightarrow3\left(t^3-27\right)-18\left(t-3\right)\ge0\)

\(\Leftrightarrow\left(t-3\right)\left(3t^2+9t+27\right)-18\left(t-3\right)\ge0\)

\(\Leftrightarrow\left(t-3\right)\left(3t^2+9t+9\right)\ge0\)

BĐT hiển nhiên đúng,do \(t\ge3\) và \(3t^2+9t+9=3\left(t+\frac{3}{2}\right)^2+\frac{9}{4}\ge\frac{9}{4}>0\)

Dấu "=" xảy ra khi t = 3 tức là \(\hept{\begin{cases}x=y=z\\xyz=1\end{cases}}\Leftrightarrow x=y=z=1\)

Chứng minh hoàn tất

Em sửa chút cho bài làm ngắn gọn hơn.

Khúc chứng minh: \(4t^4\ge18t^2+27t+81\)

\(\Leftrightarrow4t^4-18t^2-27t-81\ge0\)

\(\Leftrightarrow\left(t-3\right)\left(4t^3+12t^2+18t+27\right)\ge0\)

BĐT hiển nhiên đúng do \(t\ge3\Rightarrow\hept{\begin{cases}t-3\ge0\\4t^3+12t^2+18t+27>0\end{cases}}\)

Còn khúc sau y chang :P Lúc làm rối quá nên không nghĩ ra ạ!

Ta có

\(P=\frac{3}{4}x+\frac{1}{x}+\frac{2}{y^2}+y\)

  \(=\left(\frac{1}{x}+\frac{x}{4}\right)+\left(\frac{2}{y^2}+\frac{y}{4}+\frac{y}{4}\right)+\frac{1}{2}\left(x+y\right)\)

  \(\ge2\sqrt{\frac{1}{x}.\frac{x}{4}}+3\sqrt[3]{\frac{2}{y^2}.\frac{y}{4}.\frac{y}{4}}+\frac{1}{2}.4=1+\frac{3}{2}+2=\frac{9}{2}\)

Vậy MInP=9/2 khi \(\hept{\begin{cases}\frac{1}{x}=\frac{x}{4}\\\frac{2}{y^2}=\frac{y}{4}\\x+y=4\end{cases}\Rightarrow}x=y=2\)

Ta có \(\left(x+y\right)xy=x^2-xy+y^2\)

=> \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}\)

MÀ \(\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)^2,\frac{1}{xy}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^{^2}\)

=> \(\frac{1}{x}+\frac{1}{y}\le4\)

\(A=\frac{1}{x^3}+\frac{1}{y^3}=\frac{x^3+y^3}{x^3y^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^2\le16\)

Vậy MaxA=16 khi x=y=1/2

13 - 2\(\sqrt{2x}\) > 1

-2\(\sqrt{2x}\) > 1 - 13

-2\(\sqrt{2x}\) > -12

(-2\(\sqrt{2x}\))(-1) < (-12)(-1)

2\(\sqrt{2x}\) < 12

(2\(\sqrt{2x}\))² < (12)²

8x < 12²

8x < 144

x < 18

x < 18, x > 0

=> 0 < x < 18