A very strange algebra problem with four numbers with an average 1.

Question

Here is a Russian algebra problem from a 1999 olympiad. I can’t solve it, please help!

Find all possible values of $a,b,c,d$, if they are positive reals, their average is 1, and $$\dfrac{3-a+b(-a-ac)}{4a+4ab+4abc+4}+\dfrac{3-b+c(-b-bd)}{4b+4bc+4bcd+4}+\dfrac{3-c+d(-c-ca)}{4c+4cd+4cda+4}+\dfrac{3-d+a(-d-db)}{4d+4da+4dab+4}=0.$$

What should I use? Maybe AM-GM, i.e. that $abcd\leq1$?

Answer 1

We have $$\sum_{cyc}\frac{3-a-ab-abc}{1+a+ab+abc}=0$$ or $$\sum_{cyc}\left(\frac{3-a-ab-abc}{1+a+ab+abc}+1\right)=4$$ or $$\sum_{cyc}\frac{1}{1+a+ab+abc}=1.$$ Now, by C-S, AM-GM and Maclaurin we obtain: $$1=\sum_{cyc}\frac{1}{1+a+ab+abc}\geq\frac{16}{\sum\limits_{cyc}(1+a+ab+abc)}=$$ $$=\frac{16}{8+(a+c)(b+d)+\sum\limits_{cyc}abc}\geq\frac{16}{8+\left(\frac{a+c+b+d}{2}\right)^2+4}=1.$$ The equality occurs for $a=b=c=d=1$ only, which gives the answer.

Answer 2

Let $s$, and $t,u,v>0$ be $a+ab+abc$ (for $s$), and its cyclic cousins obtained by applying mot-a-mot the variable permutation $a\to b\to c\to d$ (and its next powers) (for $t,u,v$). Then

• because the average of $a,b,c,d$ is $1$,
• if follows that the average of $ab, bc,cd,da$ is $\le 1^2$, and
• if follows that the average of $abc, bcd,cda,dab$ is $\le 1^3$, so
• if follows that the average $W$ of $s,t,u,v$ is $\le 3$.

The given equality, $\frac 14\sum\frac {3-s}{1+s}=0$, can be rewritten as an equality between the marginal terms in the chain of inequalities: $$ \frac 14\sum\frac3{1+s} \ge \frac 3{1+W} \ge 1-\frac 1{1+W} \ge \frac 14\sum\left( 1-\frac 1{1+s}\right)\ . $$ We have used the Jensen inequalities for the convex function $s\to 3/(1+s)$, and for the concave function $s\to s/(1+s)=1-1/(1+s)$, and $W\le 3$.

The equality of the marginal terms is obtained only for equalities in between, so $W=3$, so $(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=0$, so $a=b=c=d=1$.

(All sums are cyclic sums.)