Áp dụng cosi ta có \(a.a.a.b.b\le\frac{3a^5+2b^5}{5};b.b.b.a.a\le\frac{3b^5+2a^5}{5}\)

=> \(a^5+b^5\ge a^2b^2\left(a+b\right)\)

Khi đó

\(VT\le\frac{1}{ab\sqrt{a+b}}+\frac{1}{bc\sqrt{b+c}}+\frac{1}{ac\sqrt{a+c}}\)

Áp dụng BĐT buniacoxki  ta có :

\((\frac{1}{ab\sqrt{a+b}}+\frac{1}{bc\sqrt{b+c}}+\frac{1}{ac\sqrt{a+c}})^2\le\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\left(\frac{1}{b^2\left(a+b\right)}+\frac{1}{c^2\left(b+c\right)}+...\right)\)

Mà 1/a^2+1/b^2+1/c^2=1(giả thiết)

=> \(VT\le VP\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=can(3)

hay quá 

1) ĐK: \(x\ge-1\)

\(\sqrt{9x^2+9x+4}>9x+3-\sqrt{x+1}\)

<=> \(\sqrt{9x^2+9x+4}+\sqrt{x+1}>9x+3\)(1)

TH1: 9x + 3 \(\le\)0 <=> x\(\le-\frac{1}{3}\)

(1) luôn đúng 

Th2: x\(>-\frac{1}{3}\)

<=> \(\left(\frac{1}{2}x+1-\sqrt{x+1}\right)+\left(\frac{17}{2}x+2-\sqrt{9x^2+9x+4}\right)< 0\)

<=> \(\frac{\frac{1}{4}x^2}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{\frac{253}{4}x^2}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}< 0\)

<=> \(\frac{x^2}{4}\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)< 0\)vô nghiệm 

 Vì với x \(>-\frac{1}{3}\): 

ta có: \(\frac{1}{2}x+1+\sqrt{x+1}>0\)

\(\frac{17}{2}x+2+\sqrt{9x^2+9x+4}=\frac{17}{2}x+2+\sqrt{3\left(x+\frac{1}{2}\right)^2+\frac{7}{4}}>\frac{17}{2}x+2+1>0\)

=> \(\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)>0\)với x \(>-\frac{1}{3}\) và \(x^2\ge0\)với mọi x

=> \(\frac{x^2}{4}\left(\frac{1}{\frac{1}{2}x+1+\sqrt{x+1}}+\frac{253}{\frac{17}{2}x+2+\sqrt{9x^2+9x+4}}\right)\ge0\)với x\(>-\frac{1}{3}\)

Vậy \(x< -\frac{1}{3}\)

Xin lỗi bạn kết luận bài 1 là:

\(-1\le x\le-\frac{1}{3}\)

Bài 2)  \(2+\sqrt{x+2}-x\sqrt{x+2}=x\left(\sqrt{x+2}-x\right)\)(2)

ĐK: \(x\ge-2\)

(2) <=> \(2+\sqrt{x+2}+x^2-2x\sqrt{x+2}=0\)

<=> \(8+4\sqrt{x+2}+4x^2-8x\sqrt{x+2}=0\)

<=> \(\left(2x-1\right)^2-4\left(2x-1\right)\sqrt{x+2}+4\left(x+2\right)-1=0\)

<=> \(\left(2x-1-2\sqrt{x+2}\right)^2-1=0\)

<=> \(\left(x-1-\sqrt{x+2}\right)\left(x-\sqrt{x+2}\right)=0\)

<=> \(\orbr{\begin{cases}x-1=\sqrt{x+2}\left(3\right)\\x=\sqrt{x+2}\left(4\right)\end{cases}}\)

(3) <=> \(\hept{\begin{cases}x\ge1\\x^2-3x-1=0\end{cases}}\Leftrightarrow x=\frac{3+\sqrt{13}}{2}\left(tm\right)\)

(4) <=> \(\hept{\begin{cases}x\ge0\\x^2-x-2=0\end{cases}\Leftrightarrow}x=2\left(tm\right)\)

Kết luận:...

Cái này nãy tui mới làm ở bên h_ọ_c_24 ý.

\(x\left(x-1\right)^2\ge4-x\)

\(\Leftrightarrow x\left(x^2-2x+1\right)\ge4-x\)

\(\Leftrightarrow x^3-2x^2+x\ge4-x\)

\(\Leftrightarrow x^3-2x^2+2x-4\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2\right)\ge0\)

\(\Leftrightarrow x-2\ge0\left(Vì:x^2+2>0\forall x\right)\)

\(\Leftrightarrow x\ge2\)

Vậy \(S=\left\{2;+\infty\right\}\)

@ Băng Băng @ Mình không kí hiệu tập nghiệm như vậy nhé em:

S = [ 2; \(+\infty\))

Ta có: \(2|x+1|-\left(x+4\right)>0\)

\(\Leftrightarrow2|x+1|>x+4\)

​\(\Leftrightarrow\)

• \(x+4< 0\Leftrightarrow x< -4\)
• \(x+4\ge0\Leftrightarrow x\ge-4\)
• \(2\left(x+1\right)< -\left(x+4\right)\Leftrightarrow x< -2\)
• \(2\left(x+1\right)>x+4\Leftrightarrow x>2\)

Từ trên: \(\Leftrightarrow\hept{\begin{cases}x< -4\\-4\le x\le-2\\x>2\end{cases}}\)

Qua trên ta suy ra được: \(x\in\left(-\infty;-2\right)\) hợp \(\left(2,+\infty\right)\)