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nirvanacs · 16-02-2014, 11:59 PM

Let $a,b,c\ge 0$,and such $$a+b+c=3$$
show that
$$\dfrac{a}{\sqrt{b^2+b+1}}+\dfrac{b}{\sqrt{c^2+c+ 1}}+\dfrac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}$$

Thank you

16-02-2014, 11:59 PM	#1
nirvanacs +Thành Viên+ Tham gia ngày: Dec 2010 Bài gởi: 20 Thanks: 0 Thanked 6 Times in 4 Posts	How prove this inequality a+b+c=3 Let $a,b,c\ge 0$,and such $$a+b+c=3$$ show that $$\dfrac{a}{\sqrt{b^2+b+1}}+\dfrac{b}{\sqrt{c^2+c+ 1}}+\dfrac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}$$ Thank you