a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b

Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)

Dấu = xảy ra <=> a=b=1

b) Áp dụng BĐT bunhiacopxki có:

\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)

c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)

Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) vào S ta được:

\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)

Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)

\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)

\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)

Lời giải:
$P=\frac{a^2b^2+b^2c^2+c^2a^2}{abc}$

Áp dụng BĐT AM-GM, dạng $(x+y+z)^2\geq 3(xy+yz+xz)$ ta có:

$(a^2b^2+b^2c^2+c^2a^2)^2\geq 3(a^2b^4c^2+a^4b^2c^2+a^2b^2c^4)$

$=3a^2b^2c^2(a^2+b^2+c^2)=3a^2b^2c^2$

$\Rightarrow a^2b^2+b^2c^2+c^2a^2\geq \sqrt{3}abc$

$\Rightarrow P=\frac{a^2b^2+b^2c^2+c^2a^2}{abc}\geq \sqrt{3}$

Vậy $P_{\min}=\sqrt{3}$. Giá trị này đạt tại $a=b=c=\frac{1}{\sqrt{3}}$

\(A=a^2+\dfrac{1}{16a^2}+b^2+\dfrac{1}{16b^2}+\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)

\(A\ge2\sqrt{\dfrac{a^2}{16a^2}}+2\sqrt{\dfrac{b^2}{16b^2}}+\dfrac{15}{32}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)

\(A\ge1+\dfrac{15}{32}\left(\dfrac{4}{a+b}\right)^2\ge1+\dfrac{15}{32}.4\)

\(A=\dfrac{1}{a}+\dfrac{1}{b}-\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)=\dfrac{1-a+b}{b}+\dfrac{1-b+a}{a}\)

Vì \(a^2+b^2=1\) và \(a,b>0\Leftrightarrow0< a< 1;0< b< 1\Leftrightarrow1+a-b>0;1-b+a>0\)

\(\Leftrightarrow A\ge2\sqrt{\dfrac{\left(1-a+b\right)\left(1-b+a\right)}{ab}}=2\sqrt{\dfrac{1-a^2-b^2+2ab}{ab}}=2\sqrt{2}\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\dfrac{1-a+b}{b}=\dfrac{1-b+a}{a}\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{\sqrt{2}}\)

\(A=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Áp dụng bđt Cauchy-Schwarz dạng Engel có:

\(A\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{2}}\ge\dfrac{4}{1}+\dfrac{1}{\dfrac{1}{2}}=6\)

Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)

Vậy GTNN của A=6

Thiếu đề ko e?

Đề bài nên là $a,b>0$ sao cho $a+b=1$

Lời giải:

Áp dụng BĐT  AM-GM:

$1=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}$
\(M=\frac{a^2+b^2}{ab}+ab=\frac{(a+b)^2-2ab}{ab}+ab=\frac{1}{ab}+ab-2\)

Tiếp tục áp dụng BĐT AM-GM:

\(ab+\frac{1}{16ab}\geq \frac{1}{2}\)

\(\frac{15}{16ab}\geq \frac{15}{16.\frac{1}{4}}=\frac{15}{4}\)

$\Rightarrow ab+\frac{1}{ab}\geq \frac{17}{4}$

$\Rightarrow M\geq \frac{9}{4}$

Vậy $M_{\min}=\frac{9}{4}$ khi $a=b=\frac{1}{2}$

Áp dụng bất đẳng thức Cosi ta có :

\(\frac{a}{b+c} + \frac{b+c}{4a} \geq 1;\frac{b}{c+a} + \frac{c+a}{4b} \geq 1;\frac{c}{a+b} + \frac{a+b}{4c} \geq 1\)

\(\frac{b}{a}+\frac{a}{b} \geq 2;\frac{c}{b}+\frac{b}{c} \geq 2;\frac{a}{c}+\frac{c}{a} \geq 2\)

\(\Rightarrow A=( \frac{a}{b+c} + \frac{b+c}{4a} +\frac{b}{c+a} + \frac{c+a}{4b} +\frac{c}{a+b} + \frac{a+b}{4c}) +\frac{3}{4}(\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c} +\frac{a}{c}+\frac{c}{a} )\)

\(\geq 1+1+1+\frac{3}{4} (2+2+2)=\frac{15}{2}\)

Dấu = xảy ra khi và chỉ khi a=b=c>0