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

\(\Rightarrow4=\left(a-\dfrac{1}{a}\right)^2+\left(a+\dfrac{b}{2}\right)^2-ab+2\)

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

\(M_{min}=-2\) khi \(\left\{{}\begin{matrix}a-\dfrac{1}{a}=0\\a+\dfrac{b}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left(1;-2\right);\left(-1;2\right)\)

Hi vọng là tìm GTLN:

Không mất tính tổng quát, giả sử b, c cùng phía với 1 \(\Rightarrow\left(b-1\right)\left(c-1\right)\ge0\Leftrightarrow bc\ge b+c-1\).

Áp dụng bất đẳng thức AM - GM ta có: 

\(4=a^2+b^2+c^2+abc\ge a^2+2bc+abc\Leftrightarrow2bc+abc\le4-a^2\Leftrightarrow bc\left(a+2\right)\le\left(2-a\right)\left(a+2\right)\Leftrightarrow bc+a\le2\)

\(\Rightarrow a+b+c\le3\).

Áp dụng bất đẳng thức Schwarz ta có:

\(P\le\dfrac{ab}{9}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)+\dfrac{bc}{9}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)+\dfrac{ca}{9}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)=\dfrac{1}{9}.3\left(a+b+c\right)=\dfrac{1}{3}\left(a+b+c\right)\le1\).

Đẳng thức xảy ra khi a = b = c = 1.

Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{​​}\text{​​}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)

Ta thấy: \(\text{​​}\text{​​}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)

             \(\text{​​}\text{​​}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)

Do đó: P \(\ge2+4+5=11\)

Vậy: P(min)=11  khi:  \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)

Theo đề ra, ta có:

\(a^2+b^2+c^2\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)

\(=a^3+b^3+c^3+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)

Theo BĐT Cô-si:

\(\left\{{}\begin{matrix}a^3+ab^2\ge2a^2b\\b^3+bc^2\ge2b^2c\\c^3+ca^2\ge2c^2a\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)

Do vậy \(M\ge14\left(a^2+b^2+c^2\right)+\dfrac{3\left(ab+bc+ac\right)}{a^2+b^2+c^2}\)

Ta đặt \(a^2+b^2+c^2=k\)

Luôn có \(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2=1\)

Vì thế nên \(k\ge\dfrac{1}{3}\)

Khi đấy:

\(M\ge14k+\dfrac{3\left(1-k\right)}{2k}=\dfrac{k}{2}+\dfrac{27k}{2}+\dfrac{3}{2k}-\dfrac{3}{2}\ge\dfrac{1}{3}.\dfrac{1}{2}+2\sqrt{\dfrac{27k}{2}.\dfrac{3}{2k}}-\dfrac{3}{2}=\dfrac{23}{3}\)

\(\Rightarrow Min_M=\dfrac{23}{3}\Leftrightarrow a=b=c=\dfrac{1}{3}\).

Chắc chắn đây không phải là 1 đề bài chính xác

\(A=\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(ab+\dfrac{16}{ab}\right)+\dfrac{17}{2ab}\)

\(A\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{16ab}{ab}}+\dfrac{17}{\dfrac{2\left(a+b\right)^2}{4}}\)

\(A\ge\dfrac{4}{\left(a+b\right)^2}+8+\dfrac{34}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}+8+\dfrac{34}{4^2}=\dfrac{83}{8}\)

Dấu "=" xảy ra khi \(a=b=2\)

Khúc đầu là: \(\dfrac{1}{a^4+b^2+2b^2}\) hay \(\dfrac{1}{a^4+b^2+2ab^2}\) ??

\(2a^2b\) không phải \(2ab^2\)