# How can I prove that the following function is always positive analytically?

This question is for an independent research and I have broken down the main proof to this basic sub-proof.

I want to show that the following expression is always positive. $$a^3 (52 + a (37 + a (-33 + a (29 + 2 a (11 + 4 (-2 + a) a))))) - 2 (2 + a - a^2 + a^3)^2 \delta^3$$

with the constraint $$1>a>\delta>0$$

I know this is always positive because Mathematica cannot find any instance where this is not true and I plotted this function, given the condition $$1>a>\delta>0$$, and it is always positive.

What I did was rearrange the blocks and had on the LHS $$\frac{a^3}{\delta^3}$$ which is always greater than 1 (because $$a>\delta$$). On the RHS I then have $$\frac{2 (2 + a - a^2 + a^3)^2}{(52 + a (37 + a (-33 + a (29 + 2 a (11 + 4 (-2 + a) a)))))}$$ I now attempt to show that the maximum value of the term on the RHS is less than 1 (Again I checked by plotting the function that the maximum value is around 0.185 and it is a continuously increasing function of $$a$$). However, the first order derivative is very complicated -- I took second - third and fourth order derivative and the expressions got simpler in the hope of backtracking by showing that the derivatives are all increasing in $$a$$ and minimum value is positive. However, this doesn't always work.

• You can try write $\text{RHS} - 1$ as a fraction, and prove that neither numerator nor denominator has real roots using Sturm's method. A lot of calculations but probably doable. Oct 3 at 9:18
If you just multiply this out, you get $$f(a,\delta) := 8a^9 - 16a^8 + 22a^7 - 2a^6\delta^3 + 29a^6 + 4a^5\delta^3 - 33a^5 - 6a^4\delta^3 + 37a^4 - 4a^3\delta^3 + 52a^3 + 6a^2\delta^3 - 8a\delta^3 - 8\delta^3.$$ Using that $$a > \delta > 0$$, yields \begin{align*} f(a,\delta) &> 6a^9 - 16a^8 + 16a^7 + 25a^6 - 33a^5 + 29a^4 + 44a^3\\ &= a^7(6a^2 - 16a + 16) + a^4(25a^2 - 33a + 29) + 44 \end{align*} That $$6a^2 - 16a + 16$$ and $$25a^2 - 33a + 29$$ are always positive you can easily check directly.
• Thank you so much! I am trying to figure out your intuition -- please let me know if I am wrong. So, essentially you tried to separate out the $\delta$ terms and paired them with corresponding $a$ terms as they are positive because of the constraint. Then you simply removed them and wrote the inequality and everything else follows. If you do not mind me asking, how can I develop this intuition? Is it just practice or solving similar problems? Thanks. Oct 3 at 10:07
• @user1102664 Yes, essentially. Getting rid of one variable is always a good idea. Here it was pretty easy due to $a > \delta > 0$, which means we can always estimate by $a$ or $0$ depending on the sign. And then there are not many negative signs left, so all we have to do is to pair them with sufficiently many positive terms to get something positive. Oct 3 at 10:47