\(3=a+b+c\ge3\sqrt[3]{abc}\Rightarrow abc\le1\)

BĐT tương đương:

\(3\left(ab+bc+ca\right)\ge abc\left[\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+6\right]\)

\(\Leftrightarrow3\left(ab+bc+ca\right)\ge abc\left[15-2\left(ab+bc+ca\right)\right]\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(2abc+3\right)\ge15abc\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\left(2abc+3\right)^2\ge225\left(abc\right)^2\)

Do \(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)=9abc\)

Nên ta chỉ cần chứng minh:

\(\left(2abc+3\right)^2\ge25abc\)

\(\Leftrightarrow\left(1-abc\right)\left(9-4abc\right)\ge0\) (luôn đúng với \(0< abc\le1\))

Dấu "=" xảy ra khi \(a=b=c=1\)

Từ giả thiết \(-2\le a,b,c\le3\) suy ra:

\(\left\{{}\begin{matrix}\left(a+2\right)\left(a-3\right)\le0\\\left(b+2\right)\left(b-3\right)\le0\\\left(c+2\right)\left(c-3\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a-6\le0\\b^2-b-6\le0\\c^2-c-6\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\ge a^2-6\\b\ge b^2-6\\c\ge c^2-6\end{matrix}\right.\)

\(\Rightarrow M=a+b+c\ge\left(a^2+b^2+c^2\right)-18=4\)

\(min=4\Leftrightarrow\left(a;b;c\right)=\left(2;3;3\right)\) và các hoán vị

Nhầm

\(\left(a;b;c\right)=\left(-2;3;3\right)\) và các hoán vị

đề chiều nay e thi đấy

thoi nghỉ ik e :))

Lời giải:
\(\text{VT}=\sum \frac{a^2}{a+2b^3}=\sum (a-\frac{2ab^3}{a+2b^3})=3-2\sum \frac{ab^3}{a+2b^3}\)

Áp dụng BĐT AM-GM:

\(\sum \frac{ab^3}{a+2b^3}\leq \sum \frac{ab^3}{3\sqrt[3]{ab^6}}=\frac{1}{3}\sum \sqrt[3]{a^2}\leq \frac{1}{3}\sum \frac{a+a+1}{3}=\frac{1}{9}[2(a+b+c)+3]=1\)

$\Rightarrow \text{VT}\geq 3-2.1=1$. Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c=1$

Mình làm được rồi, nhưng dù sao cũng cảm ơn bạn đã trả lời :)

Áp dụng bất đẳng thức Chevbyshev cho hai bộ đơn điệu cùng chiều \(\left(\dfrac{2}{a+b},\dfrac{2}{b+c},\dfrac{2}{c+a}\right)\) và \(\left(c\left(a+b\right),a\left(b+c\right),b\left(c+a\right)\right)\) ta có \(2c+2a+2b=\dfrac{2}{a+b}.c\left(a+b\right)+\dfrac{2}{b+c}.a\left(b+c\right)+\dfrac{2}{c+a}.b\left(c+a\right)\ge\dfrac{1}{3}\left(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\right)\left(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right)=\dfrac{2}{3}\left(ab+bc+ca\right)\left(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\right)\).

Mà \(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}=a+b+c\) nên \(ab+bc+ca\le3\).

Bất đẳng thức cần chứng minh tương đương:

\(\left(\dfrac{a^2+b^2}{a+b}-\dfrac{a^2+b^2+c^2}{a+b+c}\right)+\left(\dfrac{b^2+c^2}{b+c}-\dfrac{a^2+b^2+c^2}{a+b+c}\right)+\left(\dfrac{c^2+a^2}{c+a}-\dfrac{a^2+b^2+c^2}{a+b+c}\right)\le0\)

\(\Leftrightarrow\dfrac{a^2c+b^2c-c^2a-bc^2}{\left(a+b\right)\left(a+b+c\right)}+\dfrac{b^2a+c^2a-a^2b-ca^2}{\left(b+c\right)\left(a+b+c\right)}+\dfrac{c^2b+a^2b-b^2c-ab^2}{\left(c+a\right)\left(a+b+c\right)}\le0\)

\(\Leftrightarrow\dfrac{ac\left(a-c\right)+bc\left(b-c\right)}{a+b}+\dfrac{ba\left(b-a\right)+ca\left(c-a\right)}{b+c}+\dfrac{cb\left(c-b\right)+ab\left(a-b\right)}{c+a}\le0\) (1).

Không mất tính tổng quát giả sử \(a\geq b\geq c\).

Ta có \(\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{c+a}\\ac\left(a-c\right)+bc\left(b-c\right)\ge0\end{matrix}\right.\Rightarrow\dfrac{ac\left(a-c\right)+bc\left(b-c\right)}{a+b}\le\dfrac{ac\left(a-c\right)+bc\left(b-c\right)}{c+a}\);

\(\left\{{}\begin{matrix}\dfrac{1}{b+c}\ge\dfrac{1}{c+a}\\ba\left(b-a\right)+ca\left(c-a\right)\le0\end{matrix}\right.\Rightarrow\dfrac{ba\left(b-a\right)+ca\left(c-a\right)}{b+c}\le\dfrac{ba\left(b-a\right)+ca\left(c-a\right)}{c+a}\).

Từ đó: \(\Leftrightarrow\dfrac{ac\left(a-c\right)+bc\left(b-c\right)}{a+b}+\dfrac{ba\left(b-a\right)+ca\left(c-a\right)}{b+c}+\dfrac{cb\left(c-b\right)+ab\left(a-b\right)}{c+a}\le\dfrac{ac\left(a-c\right)+bc\left(b-c\right)+ba\left(b-a\right)+ca\left(c-a\right)+cb\left(c-b\right)+ab\left(a-b\right)}{c+a}=0\).

Do đó (1) đúng hay bđt ban đầu cũng đúng. Đẳng thức xảy ra khi a = b = c.

Ta có \(\sqrt{1+a^2}+\sqrt{2a}\le\sqrt{2\left(1+a^2+2a\right)}=\sqrt{2}\left(a+1\right)\).

Tương tự \(\sqrt{1+b^2}+\sqrt{2b}\le\sqrt{2}\left(b+1\right)\); \(\sqrt{1+c^2}+\sqrt{2c}\le\sqrt{2}\left(c+1\right)\).

Lại có \(\left(2-\sqrt{2}\right)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\le\left(2-\sqrt{2}\right)\sqrt{3\left(a+b+c\right)}\le3\left(2-\sqrt{2}\right)\).

Do đó \(B\le\sqrt{2}\left(a+b+c+3\right)+3\left(2-\sqrt{2}\right)\le6\sqrt{2}+6-3\sqrt{2}=3\sqrt{2}+6\).

Dấu "=" xảy ra khi a = b = c = 1.