Lời giải:

Đặt biểu thức đã cho là $P$

Do $a+b+c=6$ nên:

$P=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}$

$2P=\frac{2ab}{2a+b}+\frac{2bc}{2b+c}+\frac{2ca}{2c+a}$

$=b-\frac{b^2}{2a+b}+c-\frac{c^2}{2b+c}+a-\frac{a^2}{2c+a}$

$=a+b+c-\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)$

Áp dụng BĐT Cauchy-Schwarz:

$\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)\geq \frac{(b+c+a)^2}{2a+b+2b+c+2c+a}=\frac{a+b+c}{3}$

Do đó: $2P\leq a+b+c-\frac{a+b+c}{3}=\frac{2}{3}(a+b+c)=\frac{2}{3}.6=4$

$\Rightarrow P\leq 2$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

Lời giải:

Đặt biểu thức đã cho là $P$

Do $a+b+c=6$ nên:

$P=\frac{ab}{2a+b}+\frac{bc}{2b+c}+\frac{ca}{2c+a}$

$2P=\frac{2ab}{2a+b}+\frac{2bc}{2b+c}+\frac{2ca}{2c+a}$

$=b-\frac{b^2}{2a+b}+c-\frac{c^2}{2b+c}+a-\frac{a^2}{2c+a}$

$=a+b+c-\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)$

Áp dụng BĐT Cauchy-Schwarz:

$\left(\frac{b^2}{2a+b}+\frac{c^2}{2b+c}+\frac{a^2}{2c+a}\right)\geq \frac{(b+c+a)^2}{2a+b+2b+c+2c+a}=\frac{a+b+c}{3}$

Do đó: $2P\leq a+b+c-\frac{a+b+c}{3}=\frac{2}{3}(a+b+c)=\frac{2}{3}.6=4$

$\Rightarrow P\leq 2$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{ab}{6+a-c}=\frac{ab}{a+b+c+a-c}=\frac{ab}{2a+b}\)

\(=\frac{ab}{a+a+b}\le\frac{1}{9}\left(\frac{ab}{a}+\frac{ab}{a}+\frac{ab}{b}\right)=\frac{1}{9}\left(2b+a\right)\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\frac{bc}{6+b-a}\le\frac{1}{9}\left(2c+b\right);\frac{ca}{6+c-b}\le\frac{1}{9}\left(2a+c\right)\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\le\frac{1}{9}\cdot3\left(a+b+c\right)=\frac{1}{3}\cdot\left(a+b+c\right)=\frac{6}{3}=2\)

Đẳng thức xảy ra khi \(a=b=c=2\)