Problem:
Let a,b,c be positive real numbers. Prove that
5a+ba3+3b3​+5b+cb3+3c3​+5c+ac3+3a3​≥32​(a2+b2+c2)
Solution:
Solution 1: Recall the following form of Cauchy-Schwarz inequality,
y1​x12​​+y2​x22​​+…+yn​xn2​​≥y1​+y2​+…+yn​(x1​+x2​+…+xn​)2​
It also follows from the Cauchy-Schwarz inequality that x12​+x22​+x32​≥x1​x2​+x2​x3​+x3​x1​. From these two inequalities, deduce that+D17
5a+ba3​+5b+cb3​+5c+ac3​​=5a2+aba4​+5b2+bcb4​+5c2+cac4​≥5(a2+b2+c2)+(ab+bc+ca)(a2+b2+c2)2​≥61​(a2+b2+c2)​
The equality holds if and only if a=b=c.
This problem and solution were suggested by Titu Andreescu.
Solution 2 Note that
0​≤(41a+83b)(a−b)2=41a3+a2b−125ab2+83b3​
which is equivalent to
(5a+b)(−a2+25b2)≤36(a3+3b3)
Hence,
5a+ba3+3b3​≥−361​a2+3625​b2
Adding this with two other analogous inequalities completes the proof.
Discovery: The solution can be discovered naturally. We start with guessing
5a+ba3+3b3​≥ta2+(32​−t)b2
and rewrite it into
(1−5t)a3−ta2b−5(32​−t)ab2+(37​+t)b3≥0
Wishing (a−b)2 to be a factor, we use synthetic division to write the left-hand side as
(a−b)2[(1−5t)a+(2−11t)b]−(31​+12t)b3
and get t=−1/36 by setting the remainder equal to 0 .
This solution was suggested by Titu Andreescu and independently by Li Zhou, Polk State College, Winter Haven, FL.
Solution 3: It is convenient to use the shorthand notation ∑cyc ​∗ to denote the sum of the three expressions obtained from ∗ by cyclically permuting the variables a,b,c. For instance,
cyc ∑​a4b=a4b+b4c+c4a
In this notation, by clearing denominators, we may rewrite the desired inequality as
0≤cyc ∑​(190a4b+35a3b2+38ab4−35a2b3−168a3bc−60a2b2c)(1)
It is tempting to attempt to prove this using Muirhead's inequality, but this fails because we are working with cyclic sums rather than symmetric sums. For instance, it is not true that
cyc∑​a4b≥cyc∑​a3b2
(e.g., take (a,b,c)=(10,7,1)) even though Muirhead's inequality does imply the corresponding inequality for symmetric sums.
One must instead keep in mind not the statement of Muirhead's inequality but its underlying intuition: one should use "less mixed" monomials to dominate "more mixed" monomials. We will see two key techniques for realizing this intuition in the following argument. (Note that the breakdown we will give is in no way unique; there is some flexibility in the choice of how to separate (1) into tractable pieces.)
We first use what one might call a "sum of squares" argument: writing down cyclic sums of manifestly nonnegative expressions in order to match a few of the terms in (1). For instance, the following inequalities are all valid:
0≤cyc ∑​84a2b(a−c)2=cyc ∑​(84a4b−168a3bc+84a2b2c)(2)
0≤cyc ∑​235​ab2(a−b)2=cyc ∑​(235​a3b2−35a2b3+235​ab4)(3)
0≤cyc ∑​235​ab2(a−c)2=cyc ∑​(235​a3b2−35a2b2c+235​ab2c2)(4)
and these completely account for the summands 35a3b2,−35a2b3,−168a3bc in (1). We would like to add (2), (3), (4), and one more true inequality to get (1); that final inequality then would have to be
0≤cyc ∑​(2177​a4b+38ab4−2253​a2b2c)(5)
This inequality does not immediately present itself as a sum of squares, so we resort to a second technique: the weighted arithmetic-geometric mean inequality. This inequality implies that for any nonnegative real numbers u,v,w adding up to 1 ,
cyc ∑​a4b=cyc ∑​(ua4b+vb4c+wc4a)≥cyc ∑​a4u+wbu+4vcv+4w
We may then deduce that
cyc ∑​a4b≥cyc ∑​a2b2c(6)
by solving the linear equations
4u+w=2,u+4v=2,v+4w=1
and discovering that the unique real solution
(u,v,w)=136​,135​,132​
consists of nonnegative real numbers. (It is not necessary to check separately that the three numbers add up to 1 , because adding the three given equations together gives 5(u+v+w)= 5.) By switching a and b, we also obtain the valid inequality
cyc ∑​ab4≥cyc ∑​a2b2c(7)
Adding 177/2 times (6) by 177/2 plus 38 times (7) then gives (5), so this inequality is also valid. As noted earlier, we may then add (5) to (2), (3), (4) to obtain the desired inequality (1).
This solution was adapted and refined by Kiran Kedlaya from several students' solutions.
The problems on this page are the property of the MAA's American Mathematics Competitions