1.

Đặt \(\sqrt[12]{a}=x\ge0\)

\(\Rightarrow VT=2^x+2^{x^3}\ge2\sqrt{2^{x+x^3}}\ge2\) (đpcm)

Dấu "=" xảy ra khi \(x=0\) hay \(a=0\)

2.

\(y=2^{x-1}+2^{3-x}\ge2\sqrt{2^{x-1+3-x}}=4\)

\(y_{min}=4\) khi \(x-1=3-x\Leftrightarrow x=2\)

(1+a)(1+b)???

à               ,(1+a)(1+b)\(\ge\frac{9}{4}\)

cái đề là \(\left(1+a\right)\left(1+b\right)=\frac{9}{4}\)

\(A=\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\frac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)

\(\left\{{}\begin{matrix}\sqrt{x}\ge0\\x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\end{matrix}\right.\) \(\Rightarrow A\ge0\)

\(\Rightarrow A_{min}=0\) khi \(x=0\)

Với \(x\ne0\Rightarrow A=\frac{1}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}\le\frac{1}{2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}-1}=\frac{1}{2-1}=1\)

\(\Rightarrow A_{max}=1\) khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Leftrightarrow x=1\)

Ta có : \(\dfrac{a+b}{2}\ge\sqrt{ab}\) (tự cm)

Lại có : \(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}\)

Áp dụng BĐT trên ta có : : \(xy\le\left(\dfrac{x+y}{2}\right)^2\)

\(\Leftrightarrow A\ge\dfrac{x+y}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{1}{\dfrac{1}{2^2}}=4\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy...

Có: A=\(\dfrac{1}{x}+\dfrac{1}{y}\) =\(\dfrac{x+y}{xy}\) =\(\dfrac{1}{xy}\) ( do x+y=1)

     Áp dụng bđt \(\dfrac{a+b}{2}\ge\sqrt{ab}\) ,dâú bằng xảy ra khi a=b, ta có:

A=\(\dfrac{1}{x}+\dfrac{1}{y}\) =\(\dfrac{1}{xy}\) ≥ \(\dfrac{2}{x+y}\) =\(\dfrac{2}{1}\) =2 ( x+y=1)

dấu bằng xảy ra khi x=y=0,5. 

c/m bđt \(\dfrac{a+b}{2}\ge\sqrt{ab}\) ⇔ a+b ≥ 2\(\sqrt{ab}\)

                                    ⇔(a+b)² ≥ 4ab 

                                     ⇔a² +b² +2ab≥ 4ab

                                      ⇔(a-b)² ≥ 0 (luôn đúng)

   dấu bằng xảy ra khi a=b.

\(\dfrac{a+b}{2}\ge\sqrt{ab}\left(\circledast\right)\\ \Leftrightarrow a+b\ge2\sqrt{ab}\\ \Leftrightarrow\left(a+b\right)^2\ge4ab\\ \Leftrightarrow a^2+2ab+b^2-4ab\ge0\\ \Leftrightarrow a^2-2ab+b^2=\left(a-b\right)^2\ge0\left(\text{luôn đúng}\right)\)

Vậy BĐT (*) được chứng minh.

\(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{1}{xy}\)

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 \(\dfrac{x+y}{2}\ge\sqrt{xy}\\ \Rightarrow\sqrt{xy}\le\dfrac{1}{2}\\ \Rightarrow xy\le\dfrac{1}{4}\\ \Rightarrow A=\dfrac{1}{xy}\ge\dfrac{1}{\dfrac{1}{4}}=4\)

Vậy GTNN của A = 4

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

\(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}\)

Theo đề bài, ta có:

 \(\dfrac{x+y}{2}\ge\sqrt{xy}\\ \Leftrightarrow\dfrac{\left(x+y\right)^2}{4}\ge xy\\ \Leftrightarrow xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1^2}{4}=\dfrac{1}{4}\\ \Leftrightarrow xy\le\dfrac{1}{4}\\ \Leftrightarrow A\le\dfrac{x+y}{xy}=\dfrac{1}{\dfrac{1}{4}}=4\)

Vậy \(A_{min}=4\Leftrightarrow x=y=\dfrac{1}{2}\)

*Chứng minh bất đẳng thức

Ta có: \(\forall a,b\ge0\) thì \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)

\(\Leftrightarrow a+b-2\sqrt{ab}\ge0\) \(\Leftrightarrow a+b\ge2\sqrt{ab}\) \(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)  (đpcm)

Ta có: \(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\forall a,b>0\)

\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\forall a,b>0\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\forall a,b>0\)

\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\forall a,b>0\)(đpcm)

\(A=\frac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x+16}{\sqrt{x}+3}\)

+ \(A=\frac{x+16}{\sqrt{x}+3}=\frac{x-9+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}\) \(=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)

\(=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\frac{25}{\sqrt{x}+3}}-6=10-6=4\)

Dấu "=" \(\Leftrightarrow\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow\sqrt{x}+3=5\Leftrightarrow x=4\)

Vậy \(A=\frac{x+16}{\sqrt{x}+3}\)

Min A = 4 \(\Leftrightarrow x=4\)