1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

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

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

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

Áp dụng BĐT Minicopski, ta có:

\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}\ge\sqrt{\left(a+b\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2}\\ \Rightarrow P\ge\sqrt{4^2+\left(\dfrac{4}{a+b}\right)^2}=\sqrt{16+\left(\dfrac{4}{4}\right)^2}=\sqrt{17}\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=2\)

Áp dụng BĐT Cô si

⇒ P≥ \(\sqrt{2\sqrt{a^2.\dfrac{1}{a^2}}}+\sqrt{2\sqrt{b^2.\dfrac{1}{b^2}}}\)

\(=\sqrt{2}+\sqrt{2}\)

\(=2\sqrt{2}\)

Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)

Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)

\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)

\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((a^2+\frac{1}{b^2})(1+4^2)\geq (a+\frac{4}{b})^2\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{1}{\sqrt{17}}(a+\frac{4}{b})\)

Hoàn toàn tương tự với những cái còn lại và cộng theo vế suy ra:

$S\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c})$

$\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{36}{a+b+c})$ theo BĐT Cauchy-Schwarz.

Áp dụng BĐT AM-GM:

\(a+b+c+\frac{9}{4(a+b+c)}\geq 3\)

\(\frac{135}{4(a+b+c)}\geq \frac{135}{4.\frac{3}{2}}=\frac{45}{2}\)

\(\Rightarrow a+b+c+\frac{36}{a+b+c}\geq \frac{51}{2}\)

\(\Rightarrow S\geq \frac{3\sqrt{17}}{2}\)

Vậy $S_{\min}=\frac{3\sqrt{17}}{2}$

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((a^2+\frac{1}{b^2})(1+4^2)\geq (a+\frac{4}{b})^2\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{1}{\sqrt{17}}(a+\frac{4}{b})\)

Hoàn toàn tương tự với những cái còn lại và cộng theo vế suy ra:

$S\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c})$

$\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{36}{a+b+c})$ theo BĐT Cauchy-Schwarz.

Áp dụng BĐT AM-GM:

\(a+b+c+\frac{9}{4(a+b+c)}\geq 3\)

\(\frac{135}{4(a+b+c)}\geq \frac{135}{4.\frac{3}{2}}=\frac{45}{2}\)

\(\Rightarrow a+b+c+\frac{36}{a+b+c}\geq \frac{51}{2}\)

\(\Rightarrow S\geq \frac{3\sqrt{17}}{2}\)

Vậy $S_{\min}=\frac{3\sqrt{17}}{2}$

cái cuối là \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\)  nha

\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)

\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tương tự:

\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)

Cộng vế:

\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

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