Prove that $\displaystyle \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} >\sqrt{a^2+ac+c^2}$,if a,b,c>0.
I'm not so sure, here it goes...
I used Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
$\displaystyle \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} >\sqrt{a^2+ac+c^2}$
$\displaystyle \sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} \geq 2\sqrt{\sqrt{(a^2-ab+b^2)(b^2-bc+c^2)}}\geq 2\sqrt{\sqrt{(a^2-2ab+b^2)(b^2-2bc+c^2)}}=2\sqrt{\sqrt{(a-b)^2(b-c)^2}}=2\sqrt{(a-b)(b-c)}>2\sqrt{\frac{(a+c)^2}{4}}=\sqrt{(a+c)^2}>\sqrt {a^2+ac+c^2}$
You could have also substituted a, b, and c with any number greater than 0, and see if the statement returned true.
Something like.
Whops sorry, remove this-
What I meant though was substitute 1 for all three variables and see if the function is true.
$\displaystyle \sqrt{1^2-1*1+1^2}+\sqrt{1^2-1*1+1^2} >\sqrt{1^2+1*1+1^2}$
All though I could be completely wrong, so sorry if I am. Please correct.
I was thinking that if we take OA=x,OB=y,OC=z with $\displaystyle \angle AOB=60,\angle BOC=60\Righarrow AB=\sqrt{a^2-ab+b^2},BC=\sqrt{b^2-bc+c^2},AC=\sqrt{a^2+ac+c^2}$ and using the inequality of triangle in ABC we got the conclusion.