Question

[Minh Hiếu]

Ai giúp em dùng đạo hàm giải bài này với ạ

Cho a,b,c>0 và ab+bc+ca=abc

Tìm min \(A=\dfrac{a}{a+bc}+\sqrt{3}\left(\dfrac{b}{b+ca}+\dfrac{c}{c+ab}\right)\)

Answer 1

[Nguyễn Việt Lâm]

Đặt \(x=\sqrt{\dfrac{a}{bc}}\) ; \(y=\sqrt{\dfrac{b}{ca}}\) ; \(z=\sqrt{\dfrac{c}{ab}}\)

\(\Rightarrow a=\dfrac{1}{yz}\) ; \(b=\dfrac{1}{zx}\) ; \(c=\dfrac{1}{xy}\)

\(\Rightarrow xy+yz+zx=1\)

Khi đó, tồn tại một tam giác ABC sao cho:

\(x=tan\dfrac{A}{2}\) ; \(y=tan\dfrac{B}{2}\) ; \(z=tan\dfrac{C}{2}\)

Thay vào bài toán:

\(A=\dfrac{x^2}{1+x^2}+\sqrt{3}\left(\dfrac{y^2}{1+y^2}+\dfrac{z^2}{1+z^2}\right)\)

\(=\dfrac{tan^2\dfrac{A}{2}}{1+tan^2\dfrac{A}{2}}+\sqrt{3}\left(\dfrac{tan^2\dfrac{B}{2}}{1+tan^2\dfrac{B}{2}}+\dfrac{tan^2\dfrac{C}{2}}{1+tan^2\dfrac{C}{2}}\right)\)

\(=sin^2\dfrac{A}{2}+\sqrt{3}\left(sin^2\dfrac{B}{2}+sin^2\dfrac{C}{2}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{2}cosA+\dfrac{\sqrt{3}}{2}\left(2-cosB-cosC\right)\)

\(=\dfrac{1+2\sqrt{3}}{2}-\dfrac{1}{2}\left(cosA+\sqrt{3}cosB+\sqrt{3}cosC\right)\)

Xét \(B=cosA+\sqrt{3}\left(cosB+cosC\right)=cosA+2\sqrt{3}cos\dfrac{B+C}{2}cos\dfrac{B-C}{2}\)

\(\le cosA+2\sqrt{3}cos\dfrac{B+C}{2}=-2sin^2\dfrac{A}{2}+2\sqrt{3}sin\dfrac{A}{2}+1\)

Xét hàm \(f\left(t\right)=-2t^2+2\sqrt{3}sint+1\) với \(t\in\left(0;1\right)\)

\(f'\left(t\right)=-4t+2\sqrt{3}=0\Rightarrow t=\dfrac{\sqrt{3}}{2}\)

\(f\left(0\right)=1\) ; \(f\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{5}{2}\) ; \(f\left(1\right)=2\sqrt{3}-1\)

\(\Rightarrow B_{max}=\dfrac{5}{2}\)

\(\Rightarrow A\ge\dfrac{1+2\sqrt{3}}{2}-\dfrac{5}{4}=\dfrac{4\sqrt{3}-3}{4}\)