A difficult minimum

Question

Positive real numbers a, b, and c satisfy that $(a+b)(b+c)(c+a)=10$, find the minimum value of the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)+24abc$$ I have tried to square it completely $$ \prod (a^2+b^2+ab)+\frac{24}{10}abc\prod (a+b)-\frac{72}{100} \prod (a+b)^2=\frac{11}{150}\sum (a-b)^{2}(b-c)^{2}(c-a)^{2}+\frac{1}{50}\sum (a^2c+ab^2-4abc+bc^2)^2+\frac{1}{50}\sum (a^2b-4abc+ac^2+b^2c)^2\ge 0 $$ but I want a nice solution

Answer 1

From your proof we see that you proved that $72$ is a minimal value.

Another way.

After homogenization need to prove that: $$\prod_{cyc}\left(a^2+ab+b^2\right)+\frac{12}{5}abc\prod_{cyc}(a+b)\geq\frac{18}{25}\prod_{cyc}(a+b)^2.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ may by negative, and $abc=w^3$.

Thus, we need to prove that: $$27\left(3u^2v^4-u^3w^3-v^6\right)+\frac{12}{5}w^3\left(9uv^2-w^3\right)\geq\frac{18}{25}\left(9uv^2-w^3\right)^2,$$ which says that we need to prove that $f(w^3)\geq0,$ where $$f(w^3)=-w^6+A\left(u,v^2\right)w^3+B\left(u,v^2\right)$$ and $A$ and $B$ are polynomials of $u$ and $v^2$.

But $f$ is a concave function and the concave function gets a minimal value for an extremal value of $w^3$, which by $uvw$ happens for equality case of two variables.

Let $b=c$.

Since for $b=c=0$ the inequality is obvious and our inequality is homogeneous, we can assume $b=c=1$, which gives $$\left(a^2-3a+1\right)^2\geq0.$$ About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Answer 2

I came up with another solution, By the Cauchy-Schwarz inequality, we have $$ \left( a^2+ab+b^2 \right) \left( b^2+bc+c^2 \right) \left( c^2+ca+a^2 \right) =\left( \left( a^2+\frac{a\left( b+c \right)}{2}+bc \right) ^2+\frac{3}{4}a^2\left( b-c \right) ^2 \right) \left( \frac{3}{4}\left( b+c \right) ^2+\frac{1}{4}\left( b-c \right) ^2 \right) $$ $$ \ge \left( \frac{\sqrt{3}}{2}\left( b+c \right) \left( a^2+\frac{a\left( b+c \right)}{2}+bc \right) +\frac{\sqrt{3}}{4}a\left( b-c \right) ^2 \right) ^2 $$ $$ =\frac{3}{4}\left( \left( a+b \right) \left( b+c \right) \left( c+a \right) -2abc \right) ^2=\frac{3}{4}\left( 10-2abc \right) ^2 $$ $$ \text{then}\ \text{LHS}\ge \frac{3}{4}\left( 10-2abc \right) ^2+24abc=3\left( abc-1 \right) ^2+72\ge 72\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$