Can you minimize $f(a,b,c,d)$ ?

Algebra Level 4

$a,b,c,d$ are non-negative real numbers such that $ab+bc+cd+da=1$

The minimum value of $f(a,b,c,d)$ $=$ $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c}$ can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers.

Find the value of $p+q$

The answer is 4.

1 solution

Souryajit Roy
May 15, 2014

Let $A=b+c+d, B=a+c+d, C=a+b+d, D=a+b+c$ . Hence, $A+B+C+D=3(a+b+c+d)$ .

Note that ${a,b,c,d}$ and ${\frac{1}{A},\frac{1}{B},\frac{1}{C},\frac{1}{D}}$ are similarly ordered.

So by Chebychev's inequality, we have $f(a,b,c,d)≥\frac{1}{4}(a^{3}+b^{3}+c^{3}+d^{3})(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+\frac{1}{D})$ .

Again, $(A+B+C+D)(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+\frac{1}{D})≥16$

So, $f(a,b,c,d)≥\frac{4(a^{3}+b^{3}+c^{3}+d^{3})}{3(a+b+c+d)}$

Again, by Chebychev's inequality $(a^{3}+b^{3}+c^{3}+d^{3})≥\frac{1}{4}(a+b+c+d)(a^{2}+b^{2}+c^{2}+d^{2})≥\frac{1}{4}(a+b+c+d)(ab+bc+cd+da)=\frac{1}{4}(a+b+c+d)$

Hence, the minimum value of $f(a,b,c,d)$ is $\frac{1}{3}$ .

In the end: $a^2+b^2+c^2+d^2\geq (a+d)(b+c)$ , how have you proved this?

Felipe Hofmann - 6 years, 10 months ago

$(a+d)(b+c)\neq ab + bc + cd + da$ , you misread.

$a^2+b^2+c^2+d^2\ge ab+bc+cd+da$ $\iff (a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2\ge 0$

This is a classic inequality, you should memorize it.

mathh mathh - 6 years, 10 months ago

Can you minimize f ( a , b , c , d ) f(a,b,c,d) f ( a , b , c , d ) ?

The answer is 4.

1 solution

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Can you minimize $f(a,b,c,d)$ ?