$a+b+c=3, a,b,c>0$, Prove that $$a^2b^2c^2\ge (3-2a)(3-2b)(3-2c)$$
My work:
From the given inequality, we can have,
$a^2b^2c^2\ge 27-18(a+b+c)+12(ab+bc+ca)-8abc$
We can also have,$abc\le \bigg(\dfrac{a+b+c}{3}\bigg)^3=1$
So, $0\ge -36+12(ab+bc+ca)$
Again, we can have, $ab+bc+ca\le b(3-b)+\dfrac{1}{b}$
Now, I have to show that, $b(3-b)+\dfrac{1}{b}\le 3$
How can I prove this now? Please help.
Here's a geometric method, just for variety.
First (as in Yiyuan Lee's solution) we consider cases: how many of the factors on the RHS are negative?
Case 3: All three are negative; then the RHS is negative and we are done.
Case 2: Exactly two are negative; wlog, $3-2a<0$ and $3-2b<0$. This is impossible because $3-2a = a+b+c-2a = -a+b+c$, so $3-2a<0$ means $b+c<a$; similarly $3-2b<0$ means $a+c<b$; adding these inequalities yields $c<0$.
Case 1: Exactly one is negative; same as case 3.
Case 0: All three are positive. Then, as in case 2, we obtain $b+c>a$ and $a+c>b$ and $a+b>c$. Therefore there is a triangle with sides $a$,$b$,$c$. Using a standard formula for the circumradius $R$ and our hypothesis that $a+b+c=3$, we get $$ R = \frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}} = \frac{abc}{\sqrt{3(3-2a)(3-2b)(3-2c)}} $$ and so the desired inequality is equivalent to $R\ge\frac1{\sqrt3}$. So we want to prove this:
Any triangle with perimeter $3$ has circumradius at least $\frac1{\sqrt3}$.
Equivalently:
Any triangle with circumradius $1$ has perimeter at most $3\sqrt3$.
Equivalently:
Of all triangles with circumradius $1$, the equilateral triangle has the greatest perimeter.
By a standard compactness argument, there exists a triangle with maximum perimeter among those with circumradius $1$; to show that any such triangle is equilateral triangle, apply the following lemma to each vertex.
Lemma. Let $P$ and $Q$ be fixed points on a circle, and let $X$ be a variable point on that circle. If $PX+XQ$ is maximized then $X$ lies on the diameter bisecting the chord $PQ$.
Proof of lemma. Let $XY$ bisect $\angle PXQ$, with $Y$ on $PQ$. Drop perpendiculars from $P$ and $Q$ to $XY$, with feet $P'$ and $Q'$ respectively.
Then $$ PX+XQ = \frac{PP'+Q'Q}{\sin(\frac12\angle PXQ)} \le \frac{PY+YQ}{\sin(\frac12\angle PXQ)} = \frac{PQ}{\sin(\frac12\angle PXQ)} $$ Note that $\angle PXQ$ does not depend on the position of $X$ (Euclid III:21), only on what side of $PQ$ it falls on*, and that we have equality exactly when $Y$, $P'$, and $Q'$ coincide, that is, $XY$ is perpendicular to $PQ$, as desired.
*It doesn't matter for us which of the two cases is the sharp one, but of course it's when $X$ is on the major arc associated with $PQ$.
I think you can check this inequality by Lagrange multiplier method. Consider the following Lagrange function: \begin{equation} L(a,b,c)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)+\lambda(a+b+c-3) \end{equation} For finding the extreme value, we have: \begin{equation} \frac{\partial L}{\partial a}=2ab^2c^2+2(3-2b)(3-2c)+\lambda=0\\ \frac{\partial L}{\partial b}=2a^2bc^2+2(3-2a)(3-2c)+\lambda=0\\ \frac{\partial L}{\partial c}=2a^2b^2c+2(3-2a)(3-2b)+\lambda=0\\ \frac{\partial L}{\partial \lambda}=a+b+c-3=0 \end{equation} Because $a,b,c$ is symmetric in systems, I only show part of the solutions and the corresponding extreme values: \begin{equation} a=1,b=1,c=1,L=0\\ a=\frac{3}{2},b=\frac{3}{2},c=0,L=0\\ a=\sqrt[3]{2},b=\sqrt[3]{2},c=3-2\sqrt[3]{2},L\approx 0.1107011764 \end{equation} So, we obtain the minimum of $L(a,b,c)$ is $0$. In summary, under the constrain $a+b+c=3$, we have $L(a,b,c)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)+\lambda(a+b+c-3)=a^2b^2c^2-(3-2a)(3-2b)(3-2c)\geq0$.
This is the inequality you want to prove.
We consider four cases for $(3- 2a),(3-2b),(3-2c)$ : The case when all of them are non-negative, the case when exactly one of them is negative, the case when exactly two of them are negative, and the case when all of them are negative.
For the first case when all of them are non-negative, notice that by the AM-GM inequality, $$\begin{align}(3 - 2a) + (3 - 2b) &\ge 2\sqrt{(3-2a)(3-2b)}\\ 6 - 2a - 2b &\ge 2\sqrt{(3-2a)(3 - 2b)}\\ 2(a + b + c) - 2a - 2b &\ge 2\sqrt{(3-2a)(3 - 2b)}\\ c &\ge \sqrt{(3-2a)(3 - 2b)}\end{align}$$
Similarly for $(3 - 2a)(3-2c)$ and $(3 - 2b)(3 - 2c)$, we get $b \ge \sqrt{(3-2a)(3-2c)}$ and $a \ge \sqrt{(3-2b)(3-2c)}$ respectively.
Multiplying all $3$ together, we get :
$$\begin{align}abc &\ge\sqrt{(3-2a)^2(3-2b)^2(3-2c)^2}\\ \implies a^2b^2c^2 &\ge (3-2a)(3-2b)(3-2c)\end{align}$$
Hence the inequality holds for all $(3 - 2a), (3 - 2b), (3 - 2c) \ge 0$, with equality at $a = b = c = 1$.
For the second case when exactly one of them is negative, we assume without a loss of generality that $3 - 2a < 0$. Then, $$\begin{align}&3 - 2a < 0\\ \implies &a > \frac{3}{2}\\ \implies &3 - b - c > \frac{3}{2}\\ \implies &b + c < \frac{3}{2}\\ \implies &(3 - 2b)(3 - 2c) > 0\\ \implies &(3 - 2a)(3 - 2b)(3 - 2c) < 0\end{align}$$ On the other hand, $a^2b^2c^2 \ge 0$, hence the inequality holds true for the case when exactly one of them is negative.
For the case when exactly two of them is negative, we see that (without a loss of generality) $(3 - 2a), (3 - 2b) < 0 \implies a, b > \frac{3}{2}\implies a + b > 3 \implies a + b + c > 3$. But this is impossible since $a + b + c = 0$, leading to a contradiction.
Similarly, we proof again by contradiction that not all of the three can be negative. If, on the contrary all the three are negative, then $a, b, c > \frac{3}{2} \implies a + b + c > \frac{9}{2}$. But this is a contradiction since $a + b + c = 3$.
Therefore, after considering all the cases, we conclude that $$a^2b^2c^2 \ge (3-2a)(3-2b)(3-2c)$$
My initial solution assumed that all of $(3-2a), (3-2b), (3-2c) \ge 0$. But clearly this is not necessarily true, so I took sometime to write out this new solution, though I'm pretty sure there are other shorter methods.
$$\Longleftrightarrow (a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \le 27a^2b^2c^2$$
Let $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive). If $x < 0$ then $$(a+b+c)^3xyz \leq 0 < 27a^2b^2c^2$$. If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes
$$64xyz(x+y+z)^3 \leq 27(x+y)^2(y+z)^2(z+x)^2$$
Note that $$9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$$ and $$(xy+yz+zx)^2 \geq 3xyz(x+y+z)$$. Combining these completes our proof!
