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Algebra Difficulty 4.5 AIME Prove it

If a,b,c,d,ea,b,c,d,e are positive numbers bounded by pp and qq, i.e, if they lie in [p,q],0<p[p,q], 0 < p, prove that
(a+b+c+d+e)(1a+1b+1c+1d+1e)25+6(pqqp)2(a+b +c +d +e)\left(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}\right) \le 25 + 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2
and determine when there is equality.

Solution

Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but xx. Then the expression on the LHS has the form (r+x)(s+1x)=(rs+1)+sx+rx(r + x)(s + \frac{1}{x}) = (rs + 1) + sx + \frac{r}{x}, where rr and ss are fixed. But this is convex. That is to say, as xx increases if first decreases, then increases. So its maximum must occur at x=px = p or x=qx = q. This is true for each variable.
Suppose all five are pp or all five are qq, then the LHS is 25, so the inequality is true and strict unless p=qp = q. If four are pp and one is qq, then the LHS is 17+4(pq+qp)17 + 4\left(\frac{p}{q} + \frac{q}{p}\right). Similarly if four are qq and one is pp. If three are pp and two are qq, then the LHS is 13+6(pq+qp)13 + 6\left(\frac{p}{q} + \frac{q}{p}\right). Similarly if three are qq and two are pp.
pq+qp2\frac{p}{q} + \frac{q}{p} \geq 2 with equality iff p=qp = q, so if p<qp < q, then three of one and two of the other gives a larger LHS than four of one and one of the other. Finally, we note that the RHS is in fact 13+6(pq+qp)13 + 6\left(\frac{p}{q} + \frac{q}{p}\right), so the inequality is true with equality iff either (1) p=qp = q or (2) three of v,w,x,y,zv, w, x, y, z are pp and two are qq or vice versa.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.