Let $a, b,c $ are positive real number satisfying $a^2+b^2+c^2+2abc=1$.
How can I prove that $a+b+c\ge \dfrac{3}{2}$ ?
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let $$a=\cos{A},b=\cos{B},c=\cos{C},A+B+C=\pi$$ because it is known $$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
then $$a+b+c=\cos{A}+\cos{B}+\cos{C}\le\dfrac{3}{2}$$
math110
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we need to show $\ge$ not $\le ,$ right? – lab bhattacharjee Jun 15 '13 at 03:06
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This must $$a+b+c\le\dfrac{3}{2}$$, – math110 Jun 15 '13 at 03:09
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e.g. $a=1, b=c=0$ has $a+b+c=1<\frac{3}{2}$. – vadim123 Jun 15 '13 at 03:35
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Are you using Jensen's inequality, that the mean of the cosine is less than or equal to the cosine of the mean, for A,B,C $\le\frac{\pi}{2}$? – Bill Kleinhans Jun 15 '13 at 04:18