Question

Cho x,y,z>0 tìm min

P=\(\dfrac{3x}{Y+Z}\)+\(\dfrac{4y}{x+z}\)+\(\dfrac{5z}{x+y}\)

Answer 1

Lời giải:

\(P=\frac{3x}{y+z}+\frac{4y}{x+z}+\frac{5z}{x+y}\)

\(\Rightarrow P+12=\frac{3x}{y+z}+3+\frac{4y}{x+z}+4+\frac{5z}{x+y}+5\)

\(=\frac{3(x+y+z)}{y+z}+\frac{4(x+y+z)}{x+z}+\frac{5(x+y+z)}{x+y}\)

\(=(x+y+z)\left(\frac{3}{y+z}+\frac{4}{x+z}+\frac{5}{x+y}\right)\)

\(\geq (x+y+z).\frac{(\sqrt{3}+\sqrt{4}+\sqrt{5})^2}{y+z+x+z+x+y}\) (BĐT Svac-xơ)

\(=\frac{(\sqrt{3}+2+\sqrt{5})^2}{2}\)

\(\Rightarrow P\geq \frac{(2+\sqrt{3}+\sqrt{5})^2}{2}-12\) (min)

Dấu bằng xảy ra khi \(\frac{\sqrt{3}}{y+z}=\frac{2}{x+z}=\frac{\sqrt{5}}{x+y}\)