Let $a,b,c$ be real numbers such that $a+b+c=5$. Prove that $$(a^2+3)(b^2+3)(c^2+3) \geq 192.$$
Proposed by Marius Stanean, Zalau, Romania
Solution:
Let $f(a,b,c)=(a^2+3)(b^2+3)(c^2+3)$ and $g(a,b,c)=a+b+c-5$. Consider the Lagrangian function
$$L(a,b,c,\lambda)=f(a,b,c)-\lambda g(a,b,c)=(a^2+3)(b^2+3)(c^2+3)-\lambda(a+b+c-5),$$
where $\lambda \in \mathbb{R}$. Observe that $f(a,b,c)$ is continuous on the compact set defined by $g(a,b,c)$, so by Weierstrass Theorem, there exists a global minimum. By the method of Lagrange Multipliers, a maximum or a minimum for $f(a,b,c)$ subject to the constraint $g(a,b,c)=0$ must be a stationary point of $L$. Therefore a maximum or a minimum satisfies
$$\renewcommand{\arraystretch}{2} \begin{array}{rcl} \dfrac{\partial L}{\partial a} & = & 0 \\ \dfrac{\partial L}{\partial b} & = & 0 \\ \dfrac{\partial L}{\partial c} & = & 0 \\ \dfrac{\partial L}{\partial \lambda} & = & 0, \end{array}$$ i.e.
$$\begin{array}{rcl} 2a(b^2+3)(c^2+3)+\lambda & = & 0 \\ 2b(a^2+3)(c^2+3)+\lambda & = & 0 \\ 2c(a^2+3)(b^2+3)+\lambda & = & 0 \\ a+b+c-5 & = & 0. \end{array}$$
Subtracting side by side the first two equations and simplifying, then doing the same for the second and the third equation and for the third and the first equation, we obtain
$$\begin{array}{rcl} (a-b)(3-ab) & = & 0 \\ (b-c)(3-bc) & = & 0 \\ (c-a)(3-ca) & = & 0 \\ a+b+c-5 & = & 0. \end{array}$$
So, up to symmetry, we get $$(a,b,c) \in \left\{\left(\dfrac{5}{3},\dfrac{5}{3},\dfrac{5}{3}\right),\left(\dfrac{3}{2},\dfrac{3}{2},2 \right),\left(1,1,3\right)\right\}.$$ Since $f\left(\dfrac{5}{3},\dfrac{5}{3},\dfrac{5}{3}\right)=\left(\dfrac{52}{9}\right)^3$, $f\left(\dfrac{3}{2},\dfrac{3}{2},2\right)=\dfrac{3087}{16}$, $f(1,1,3)=192$, we conclude that $192$ is the constrained minimum, i.e.
$$(a^2+3)(b^2+3)(c^2+3) \geq 192.$$
No comments:
Post a Comment