Question

[Duong Nguyen Tuan]

Cho 3 số thực dương a,b,c thỏa mãn \(a+b+c\le2\)

Tìm GTNN của \(P=21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+2017\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Answer 1

[Nyatmax]

Ta co:

\(P\ge21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{2017.9}{2}\)

\(=21\left(a^2+b^2+c^2\right)+12\left(a+b+c\right)^2+\frac{18153}{2}\)

\(\Leftrightarrow\frac{P}{\left(a+b+c\right)^2}\ge21\left[\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{a+b+c}\right)^2+\left(\frac{c}{a+b+c}\right)^2\right]+12+\frac{\frac{18153}{2}}{\left(a+b+c\right)^2}\)

Dat \(\left(\frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}\right)\rightarrow\left(x;y;z\right)\)

\(\Rightarrow x+y+z=1\)

\(\Rightarrow\left(a+b+c\right)^2=\frac{a^2}{x^2}\)

BDT tro thanh:

\(\frac{P}{\left(a+b+c\right)^2}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\)

\(\Leftrightarrow\frac{P}{\frac{a^2}{x^2}}\ge21\left(x^2+y^2+z^2\right)+12+\frac{18153}{2\left(a+b+c\right)^2}\ge21.\frac{\left(x+y+z\right)^2}{3}+12+\frac{18153}{8}\)

\(\Leftrightarrow\frac{x^2P}{a^2}\ge7+12+\frac{18153}{8}\)

Ta lai co:\(x=\frac{a}{a+b+c}\ge\frac{a}{2}\Rightarrow a^2\le4x^2\)

Suy ra:\(\frac{x^2P}{a^2}\ge\frac{x^2P}{4x^2}=\frac{P}{4}\)

\(\Rightarrow\frac{P}{4}\ge\frac{18503}{8}\)

\(\Leftrightarrow P\ge\frac{18503}{2}\)

Dau '=' xay ra khi \(a=b=c=\frac{2}{3}\)

Vay \(P_{min}=\frac{18503}{2}\)khi \(a=b=c=\frac{2}{3}\)