ĐKXĐ : \(x\ge2;y\ge3\)

\(\Rightarrow S=\sqrt{x-2}+\sqrt{y-3}\ge1\)

Dấu " = " xảy ra \(\Leftrightarrow\orbr{\begin{cases}x=2;y=4\\y=3;x=3\end{cases}}\)

\(S=\sqrt{x-2}+\sqrt{y-3}\)

\(\Rightarrow S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\)

\(\Rightarrow S^2=x-2+2\sqrt{\left(x-2\right)\left(y-3\right)}+y-3\)

\(\Rightarrow S^2=x+y-5+2\sqrt{\left(x-2\right)\left(y-3\right)}\)

\(\Rightarrow S^2=1+2\sqrt{\left(x-2\right)\left(y-3\right)}\)

Vì \(2\sqrt{\left(x-2\right)\left(y-3\right)}\ge0\)

\(\Rightarrow1+2\sqrt{\left(x-2\right)\left(y-3\right)}\ge1\)

\(\Rightarrow S^2\ge1\Leftrightarrow\orbr{\begin{cases}S\ge1\left(tm\right)\\S\le-1\left(ktm\right)\end{cases}}\)

\(\Rightarrow S_{min}=1\Leftrightarrow2\sqrt{\left(x-2\right)\left(y-3\right)}=0\)

TH1 : \(x-2=0\Leftrightarrow x=2\Rightarrow y=6-2=4\)

Th2 : \(y-3=0\Rightarrow y=3\Rightarrow x=6-3=3\)

Vậy \(S_{min}=1\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)hoặc \(x=y=3\)

Áp dụng bđt Bu-nhi-a-cốp-xki ta có

\(S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\le\left(1+1\right)\left(x+y-5\right)=2\left(6-5\right)=2\)(vì \(x+y=6\) )

\(\Rightarrow S^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le S\le\sqrt{2}\)

\(\Rightarrow minS=-\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{\sqrt{x-2}}{1}=\frac{\sqrt{y-3}}{1}\\x+y=6\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2,5\\y=3,5\end{cases}}\)

Lời giải:
Áp dụng BĐT AM-GM:

$x^6+\frac{1}{8}+\frac{1}{8}\geq 3\sqrt[3]{\frac{x^6}{64}}=\frac{3}{4}x^2$

$y^6+\frac{1}{8}+\frac{1}{8}\geq \frac{3}{4}y^2$

Cộng 2 BĐT trên và thu gọn theo vế thì:

$A+\frac{1}{2}\geq \frac{3}{4}(x^2+y^2)$

$\Leftrightarrow A+\frac{1}{2}\geq \frac{3}{4}$

$\Leftrightarrow A\geq \frac{1}{4}$

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Lại có:

$x^2+y^2=1\Rightarrow x^2\leq 1; y^2\leq 1\Rightarrow x^4\leq 1; y^4\leq 1$

Khi đó:

$x^6\leq x^2; y^6\leq y^2$

$\Rightarrow x^6+y^6\leq x^2+y^2$

$\Rightarrow A\leq 1$
Vậy $A_{\min}=\frac{1}{4}; A_{\max}=1$