\(\dfrac{a}{2}+\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{a+b}{5}=\dfrac{a}{5}+\dfrac{b}{5}\)

We have:

\(\dfrac{a}{2}\ge\dfrac{a}{5}\)  (Equal sign occurs  \(\Leftrightarrow a=0\))

\(\dfrac{a}{3}\ge\dfrac{a}{5}\)  (Equal sign occurs  \(\Leftrightarrow a=0\))

\(\Leftrightarrow\dfrac{a}{2}+\dfrac{a}{3}\ge\dfrac{a+b}{5}\) 

Equal sign occurs  \(\Leftrightarrow a=b=0\)

Transform: \(\overline{a,bc}\div\left(a+b+c\right)=0,25\)

\(\Leftrightarrow\overline{abc}=25\left(a+b+c\right)\Rightarrow\overline{abc}⋮25\)

\(\Rightarrow\overline{bc}=\left\{25;50;75\right\}\) (attention: kind \(\overline{bc}=00\) because \(b=c\))

+) If  \(\overline{bc}=25\Rightarrow\overline{a25}=25\left(a+7\right)\)

\(\Rightarrow100a+25=25a+175\Rightarrow75a=150\Rightarrow a=2\) (unsatisfactory)

+) If  \(\overline{bc}=50\Rightarrow\overline{a50}=25\left(a+5\right)\)

\(\Rightarrow100a+50=25a+125\Rightarrow75a=75\Rightarrow a=1\)

+) If  \(\overline{bc}=75\Rightarrow\overline{a75}=25\left(a+12\right)\)

\(\Rightarrow100a+75=25a+300\Rightarrow75a=225\Rightarrow a=3\)

Conclude:...

Without reducing generality, we assume \(a>b>c\)

\(\Rightarrow\dfrac{1}{\left[a,b\right]}=\dfrac{1}{ab}\le\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{1}{6}\)

Same: \(\left\{{}\begin{matrix}\dfrac{1}{\left[b,c\right]}=\dfrac{1}{bc}\le\dfrac{1}{15}\\\dfrac{1}{\left[a,c\right]}=\dfrac{1}{ac}\le\dfrac{1}{10}\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{1}{6}+\dfrac{1}{15}+\dfrac{1}{10}=\dfrac{1}{3}=VP\)  (The thing must prove)

Use BĐT Cauchy - Schwarz form Engel we have:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\) (Đpcm)

Equal sign occurs \(\Leftrightarrow a=b=c\)

We have BĐT equivalent:

\(\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)\) \(\ge\dfrac{9}{2}\)

\(\Leftrightarrow\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)

Use AM - GM inequality we have:

\(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{3}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)

\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (Thing must prove)

The equality occurs \(\Leftrightarrow a=b=c\)