A simple inequality question

Question

Let $a,b,c,d,e,f$ be positive real numbers such that $$a+b=c+d=e+f$$ along with the condition that $$a>c>e$$ and $$b<d<f$$ Is it possible that $c^2<ae$ and $d^2<bf$ are both satisfied simultaneously? The examples that I have generated suggest that this is not possible. However, I cannot find a proof.

Answer 1

Denote the common sum with $S$: $$ S = a+b=c+d=e+f $$ and assume that $c^2<ae$ and $d^2<bf$ are both satisfied. Then, using the inequality between geometric and arithmetic mean: $$ S = c + d < \sqrt{ae} + \sqrt{bf} \le \frac{a+e}{2} + \frac{b+f}{2} = S $$ which is a contradiction.

Note that the conditions $a>c>e$ and $b<d<f$ are not needed for the conclusion.

Answer 2

Let it be possible and $a+b=c+d=e+f=k$.

Thus, $k>a$ and $$(k-c)^2<(k-a)(k-e)$$ or $$c^2<ae+k(2c-a-e).$$

If $c\geq\frac{a+e}{2}$ then by AM-GM $$c^2<ae\leq\left(\frac{a+e}{2}\right)^2\leq c^2,$$ which is a contradiction.

It est, $c<\frac{a+e}{2}$ and $$c^2<ae+k(2c-a-e)<ae+a(2c-a-e),$$ which gives $$a^2+c^2<2ac$$ or $$(a-c)^2<0,$$ which is a contradiction again.

We got that it's impossible.

Done!