$\dfrac {1}{h_a}+\dfrac{1}{h_b}>\dfrac {1}{h_c}>| \dfrac {1}{h_a}-\dfrac{1}{h_b}|$

Question

Ucgenin yukseklikleri arasinda verilen yukaridaki esitsizligi kanitlayiniz.

Comments

ilk esitsizlik dogruysa ikincisi de direkt dogru oluyor. 

Answer 1

Alan geregi $$a\cdot h_a=b\cdot h_b=c\cdot h_c$$ saglanir. Buradan $$\frac{1}{h_a}+\frac{1}{h_b}=\frac{a}{c}\cdot \frac{1}{h_c}+\frac{b}{c}\cdot \frac{1}{h_c}=\frac{a+b}{c}\cdot \frac{1}{h_c}\ge 1\cdot \frac{1}{h_c}=\frac{1}{h_c}$$ saglanir. ($a+b\ge c$ olmasi gerektigini kullandik).

Ust formul $a,b,c$ siralamasina bagli olmadigindan $$\frac{1}{h_c}+\frac{1}{h_b}\ge \frac{1}{h_a}$$ ve $$\frac{1}{h_c}+\frac{1}{h_a}\ge \frac{1}{h_b}$$ saglanir. Yani $$\frac{1}{h_c}\ge \max\left\{ \frac{1}{h_a}-\frac{1}{h_b}, \frac{1}{h_b}-\frac{1}{h_a}  \right\}=\left|\frac{1}{h_a}-\frac{1}{h_b}\right|$$ saglanir.