Let a,b,c be positive real numbers such that a+b+c=3. Prove that a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.
Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam
Solution:
Let f(x)=x\sqrt{x+3}. Observe that f''(x)=\dfrac{3(x+4)}{4\sqrt{(x+3)^3}}>0 \qquad \forall x>0. Therefore, by Jensen's Inequality, we have 2=f(1)=f\left(\dfrac{a+b+c}{3}\right) \leq \dfrac{f(a)+f(b)+f(c)}{3}=\dfrac{a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3}}{3}, i.e. a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.
Let f(x)=x\sqrt{x+3}. Observe that f''(x)=\dfrac{3(x+4)}{4\sqrt{(x+3)^3}}>0 \qquad \forall x>0. Therefore, by Jensen's Inequality, we have 2=f(1)=f\left(\dfrac{a+b+c}{3}\right) \leq \dfrac{f(a)+f(b)+f(c)}{3}=\dfrac{a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3}}{3}, i.e. a\sqrt{a+3}+b\sqrt{b+3}+c\sqrt{c+3} \geq 6.
No comments:
Post a Comment