Question

[khoimzx]

Cho a,b,c là số đo ba cạnh của tam giác . Chứng minh rằng:

\(\frac{\sqrt{a}}{b+c-a}+\frac{\sqrt{b}}{c+a-b}+\frac{\sqrt{c}}{a+b-c}\ge\frac{a+b+c}{\sqrt{abc}}\)

Answer 1

[Nguyễn Việt Lâm]

Đặt \(\left\{{}\begin{matrix}b+c-a=x>0\\c+a-b=y>0\\a+b-c=z>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{y+z}{2}\\b=\frac{z+x}{2}\\c=\frac{x+y}{2}\end{matrix}\right.\)

BĐT trở thành: \(\frac{\sqrt{y+z}}{\sqrt{2}x}+\frac{\sqrt{z+x}}{\sqrt{2}y}+\frac{\sqrt{x+y}}{\sqrt{2}z}\ge\frac{x+y+z}{\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{8}}}\)

\(\Leftrightarrow\frac{\sqrt{y+z}}{x}+\frac{\sqrt{z+x}}{y}+\frac{\sqrt{x+y}}{z}\ge\frac{4\left(x+y+z\right)}{\sqrt{\left(x+y\right)\left(y+z\right)\left(z+x\right)}}\)

\(\Leftrightarrow\frac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}+\frac{\left(z+x\right)\sqrt{\left(y+z\right)\left(y+x\right)}}{y}+\frac{\left(x+y\right)\sqrt{\left(z+x\right)\left(z+y\right)}}{z}\ge4\left(x+y+z\right)\)

Ta có:

\(\frac{\left(y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)}}{x}\ge\frac{\left(y+z\right)\left(x+\sqrt{yz}\right)}{x}=y+z+\frac{\left(y+z\right)\sqrt{yz}}{x}\ge y+z+\frac{2yz}{x}\)

Tương tự: \(\frac{\left(z+x\right)\sqrt{\left(y+z\right)\left(y+x\right)}}{y}\ge z+x+\frac{2zx}{y}\) ; \(\frac{\left(x+y\right)\sqrt{\left(z+x\right)\left(z+y\right)}}{z}\ge x+y+\frac{2xy}{z}\)

Cộng vế với vế:

\(VT\ge2\left(x+y+z\right)+2\left(\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}\right)\ge2\left(x+y+z\right)+2\left(x+y+z\right)\)

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)