Question

Cho tam giác ABC, x, y, z ∈ R. Chứng minh:

\(\frac{cosA}{x}+\frac{cosB}{y}+\frac{cosC}{z}\) ≤ \(\frac{x^2+y^2+z^2}{2xyz}\)

Answer 1

Chỉ đúng với \(x;y;z\in R^+\)

Nói chung là ta cần chứng minh

\(x^2+y^2+z^2\ge2xycosC+2zxcosB+2yzcosA\)

\(\Leftrightarrow x^2-2x\left(ycosC+zcosB\right)+y^2+z^2-2yzcosA\ge0\)

\(\Leftrightarrow\left(x-ycosC-zcosB\right)^2-\left(ycosC+zcosB\right)^2+y^2+z^2-2yzcosA\ge0\)

\(\Leftrightarrow\left(x-ycosC-zcosB\right)^2-y^2cos^2C-z^2cos^2B+y^2+z^2-2yz\left(cosB.cosC+cosA\right)\ge0\)

\(\Leftrightarrow\left(x-ycosC-zcosB\right)^2+y^2\left(1-cos^2C\right)+z^2\left(1-cos^2B\right)-2yz\left(cosB.cosC-cos\left(B+C\right)\right)\ge0\)

\(\Leftrightarrow\left(x-ycosC-zcosB\right)^2+y^2sin^2C+z^2.sin^2B-2yz.sinB.sinC\ge0\)

\(\Leftrightarrow\left(x-ycosC-zcosB\right)^2+\left(ysinC-zsinB\right)^2\ge0\) (luôn đúng)