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Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:


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up vote 9 down vote accepted

By squaring $a+b+c=3$ we get $$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=9.$$

From the AM-GM inequality (or from the fact that $(x-y)^2=x^2+y^2-2xy\ge 0$, i.e. $2xy\le x^2+y^2$) we have $$ab+ac+bc \le \frac{a^2+b^2}2+\frac{a^2+c^2}2+\frac{b^2+c^2}2=a^2+b^2+c^2,$$ i.e. $\frac12(a^2+b^2+c^2) \ge \frac12(ab+ac+bc)$, which is equivalent to $\frac12(a^2+b^2+c^2) - \frac12(ab+ac+bc) \ge0$.

By adding the above equality and inequality together you get $$\frac32(a^2+b^2+c^2+ab+ac+bc)\ge9,$$ which is equivalent to $$a^2+b^2+c^2+ab+ac+bc\ge6.$$

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By the Cauchy-Schwarz inequality, $$6=1*(a+b)+1*(b+c)+1*(a+c)\leq\sqrt{1^2+1^2+1^2}\sqrt{(a+b)^2+(b+c)^2+(a+c)^2}$$ In other words $$(a+b)^2+(b+c)^2+(a+c)^2\geq 12$$ which is the same as the desired inequality.

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If $x,y,z$ are nonnegative reals, then $x^2+y^2+z^2\ge xy+yz+zx$ (with equality iff $x=y=z$), hence $3(x^2+y^2+z^2)\ge x(x+y+z)+y(y+z+x)+z(z+x+y) = (x+y+z)^2$ (with equality iff $x=y=z$). Letting $x=a+b, y=b+c, z=a+c$, we find $x+y+z=6$ and $3(a+b)^2+3(b+c)^2+3(c+a)^2 \ge 36$. Note that $(a+b)^2+(b+c)^2+(c+a)^2= 2(a^2+b^2+c^2+ab+bc+ca)$ so that we actually showed $$ a^2+b^2+c^2+ab+bc+ca\ge 6$$ with equality iff $a=b=c$.

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$$ a^2 + b^2 + c^2 + ab + bc + ac = (a+b+c)^2 - (ab + bc + ac) = 9 - (ab + bc + ac)$$ Now it remains to show that max value of $(ab + bc + ac)$ is $3$. For that, we know the AM-GM equality ( for $a,b, c >0$ ) that $3(a^2 + b^2 + c^2) \geq (a+ b + c)^2 \geq 3(ab +bc +ac)$. From the last two part we have $(a+ b + c)^2 \geq 3(ab +bc +ac) \implies 9 \geq 3 (ab +bc +ac) \implies 3 \geq ab +bc +ac$

Hence we have $$ a^2 + b^2 + c^2 + ab + bc + ac = 9 - (ab + bc + ac) \geq 9 - 3 = 6$$

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