\(\left(\sqrt{4x+1}+\sqrt{4y+1}+\sqrt{4z+1}\right)^2\le\left(1^2+1^2+1^2\right)\left(4x+1+4y+1+4z+1\right)=21.\)

\(\Leftrightarrow\sqrt{4x+1}+\sqrt{4y+1}+\sqrt{4z+1}\le\sqrt{21}\left(đpcm\right)\)

Dấu "=" xra :

\(\frac{4x+1}{1}=\frac{4y+1}{1}=\frac{4z+1}{1}\Rightarrow x=y=z=\frac{1}{3}\)

\(\frac{\sqrt{x-1}.\sqrt{2x+2}}{\sqrt{x^2-1}}=\frac{\sqrt{x-1}.\sqrt{2\left(x+1\right)}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\)

\(=\frac{\sqrt{x-1}.\sqrt{2}.\sqrt{x+1}}{\sqrt{x-1}.\sqrt{x+1}}\)

\(=\frac{\sqrt{2}}{1}=\sqrt{2}\)

\(VT\le\frac{1}{\sqrt[3]{9}}\left(\frac{a+2b+3+3}{3}+\frac{b+2c+3+3}{3}+\frac{c+2a+3+3}{3}\right)\)

\(=\frac{1}{\sqrt[3]{9}}.\frac{3\left(a+b+c\right)+18}{3}=\frac{9}{\sqrt[3]{9}}=\sqrt[3]{81}=3\sqrt[3]{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)

\(a^3+b^3\ge\left(a+b\right)ab\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-\left(a+b\right)ab\ge0\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0.\)(Lđ với mọi a,b >0)

Dấu "=" xra khi a=b

\(a^3+b^3\ge ab\left(a-b\right)\)

\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)

\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2-ab\right)\ge0\)

\(\Rightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)

Vì \(a,b>0\Rightarrow a+b>0\)và \(\left(a-b\right)^2\ge0\)

\(\Rightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)( luôn đúng )

\(\Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)\(\left(đpcm\right)\)

#)Giải :

a)\(\Delta ABC\)vuông tại A (gt) \(\Rightarrow\widehat{BAD}+\widehat{DAC}=90^o\left(1\right)\)

\(\Delta HAD\)vuông tại H (gt)\(\Rightarrow\widehat{HDA}+\widehat{HAD}=90^o\left(2\right)\)

Vì AD là tia phân giác của \(\widehat{HAC}\Rightarrow\)\(\widehat{HAD}=\widehat{DAC}\left(3\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\widehat{BAD}=\widehat{DAC}\)

\(\Rightarrow\Delta ABD\)cân tại A

b) Từ cmt \(\Rightarrow AB=BD\)(tính chất của tam giác cân)

Đặt \(AB=BD=x\)

Áp dụng hệ thức lượng trong tam giác vuông ABC 

\(\Rightarrow AB^2=HB.HC\)

Hay \(x^2=\left(x-6\right)25\)

\(\Rightarrow x^2-25+150=0\)

\(\Rightarrow\left(x-10\right)\left(x-15\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-10=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=10\\x=15\end{cases}}}\)

Vậy AB = 10 hoặc AB = 15