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Algebra Difficulty 6.2 National olympiad Prove it

Proposition 3 Let a,b,ca, b, c be the side lengths of triangle ABCABC, Δ\Delta its area, and λ,u,v\lambda, u, v any real numbers, then
(uv+vλ+λu)abc⩾4λuv(λa2+ub2+vc2)Δ(u v+v \lambda+\lambda u) a b c \geqslant 4 \sqrt{\lambda u v\left(\lambda a^{2}+u b^{2}+v c^{2}\right)} \Delta

Equality in (4) holds if and only if λa2(−a2+b2+c2)=ub2(a2−b2+c2)=vc2(a2+b2−c2)\lambda a^{2}\left(-a^{2}+b^{2}+c^{2}\right)=u b^{2}\left(a^{2}-b^{2}+c^{2}\right)=v c^{2}\left(a^{2}+b^{2}-c^{2}\right).

Solution

To prove in the theorem, let x′=uva2,y′=vλb2,z′=λuc2,x=−a2+b2+c2,y=x^{\prime}=\frac{u v}{a^{2}}, y^{\prime}=\frac{v \lambda}{b^{2}}, z^{\prime}=\frac{\lambda u}{c^{2}}, x=-a^{2}+b^{2}+c^{2}, y= a2−b2+c2,z=a2+b2−c2a^{2}-b^{2}+c^{2}, z=a^{2}+b^{2}-c^{2}, then
(y′z′+z′x′+x′y′)=4Δ\sqrt{\left(y^{\prime} z^{\prime}+z^{\prime} x^{\prime}+x^{\prime} y^{\prime}\right)}=4 \Delta

Substituting into equation (4), we can obtain equation (4).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.