Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 .$Prove that

navi_09220114 wrote: Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 $, find the largest positive real $ k $ so that for all $a,b,c,$ we have $$ a^2+b^2+c^2+k(abc-1)\ge 3 $$ sqing wrote: Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 .$Prove that$$ a^2+b^2+c^2+\frac{3}{2}(abc-1)\ge 3\geq a^2+b^2+c^2+6(abc-1) $$

This is much better ]Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 .$Prove that$$ a^2+b^2+c^2+\frac{3}{2}(abc-1)\ge 3\geq a^2+b^2+c^2+6(\sqrt{abc}-1).$$

Is there a solution to the other inequalities shown here?

sqing wrote: Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 .$Prove that $$ a^2+b^2+c^2+\frac{3}{2}(abc-1)\ge 3\geq a^2+b^2+c^2+6(abc-1) $$

navi_09220114 wrote: Is there a solution to the other inequalities shown here? no

navi_09220114 wrote: Is there a solution to the other inequalities shown here? For my inequality there is not only solution, but also a generalization, which will soon be published and so, I'll soon post it here

We claim $k=\frac 32$. We first show that it works. We have to show \[a^2+b^2+c^2+\frac 32abc\geq\frac 92\]Schur's Inequality tells us that \[a^3+b^3+c^3+3abc\geq a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\]which rewrites to \[2(a+b+c)(a^2+b^2+c^2)+9abc\geq (a+b+c)^3\]after which substituting $a+b+c=3$ and dividing by $6$ gives the original inequality. Now, to show that $k\leq\frac 32$, plug in $a=0$. We get \[b^2+(3-b)^2\geq k+3\]However, over the interval $[0,3]$, the minimum of that function is $1.5^2\cdot 2=4.5$, implying $k\leq\frac 32$ as desired.

sqing wrote: Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 .$Prove that$$ ab+bc+ca+\frac{3}{2}(abc-1)\ge 3\geq a^2+b^2+c^2+6(\sqrt{abc}-1) $$ The first half was proved in the above post. To prove the second half, we know \[3abc(a+b+c)\leq (ab+bc+ca)^2\](apply $(x+y+z)^2\geq 3(xy+yz+zx)$ to $x=ab,y=bc,z=ca$) so thus \[a^2+b^2+c^2+6\sqrt{abc}\leq a^2+b^2+c^2+2(ab+bc+ca)=(a+b+c)^2=9\]

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sqing wrote: Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 .$Prove that$$ ab+bc+ca+\frac{3}{2}(abc-1)\ge 3...$$ It's of course, wrong. Try $a=3,b=c=0.$

sqing wrote: Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 .$Prove that$$ \frac{1}{2}(a^2+b^2+c^2+ab+bc+ca)+\frac{3}{2}(abc-1)\ge 3\geq a^2+b^2+c^2+6(\sqrt{abc}-1) $$ The right inequality belongs to me, so it is true. But the left is not mine and it is false. Try $a=b$ and $c=0.$

I'm looking forward for a correct left side inequality

For example, this is correct Let $a, b, c,$ be nonnegative reals with $ a+b+c=3 .$Prove that$$ \frac{1}{2}(a^2+b^2+c^2+ab+bc+ca)+\frac{3}{8}(abc-1)\ge 3\geq a^2+b^2+c^2+6(\sqrt{abc}-1).$$

any proofs to all of the claimed inequalities here?