Problem:
Let a,b,c be positive real numbers. Prove that
2a2+(b+c)2(2a+b+c)2+2b2+(c+a)2(2b+c+a)2+2c2+(a+b)2(2c+a+b)2≤8
Solution:
Let a,b,c be positive real numbers. Prove that
2a2+(b+c)2(2a+b+c)2+2b2+(c+a)2(2b+c+a)2+2c2+(a+b)2(2c+a+b)2≤8
First Solution: By multiplying a,b, and c by a suitable factor, we reduce the problem to the case when a+b+c=3. The desired inequality reads
2a2+(3−a)2(a+3)2+2b2+(3−b)2(b+3)2+2c2+(3−c)2(c+3)2≤8
Set
f(x)=2x2+(3−x)2(x+3)2
It suffices to prove that f(a)+f(b)+f(c)≤8. Note that
f(x)=3(x2−2x+3)x2+6x+9=31⋅x2−2x+3x2+6x+9=31(1+x2−2x+38x+6)=31(1+(x−1)2+28x+6)≤31(1+28x+6)=31(4x+4).
Hence,
f(a)+f(b)+f(c)≤31(4a+4+4b+4+4c+4)=8
as desired.
Second Solution: Note that
(2x+y)2+2(x−y)2=4x2+4xy+y2+2x2−4xy+2y2=3(2x2+y2)
Setting x=a and y=b+c yields
(2a+b+c)2+2(a−b−c)2=3(2a2+(b+c)2)
Thus, we have
2a2+(b+c)2(2a+b+c)2=2a2+(b+c)23(2a2+(b+c)2)−2(a−b−c)2=3−2a2+(b+c)22(a−b−c)2
and its analogous forms. Thus, the desired inequality is equivalent to
2a2+(b+c)2(a−b−c)2+2b2+(c+a)2(b−a−c)2+2c2+(a+b)2(c−a−b)2≥21
Because (b+c)2≤2(b2+c2), we have 2a2+(b+c)2≤2(a2+b2+c2) and its analogous forms. It suffices to show that
2(a2+b2+c2)(a−b−c)2+2(a2+b2+c2)(b−a−c)2+2(a2+b2+c2)(c−a−b)2≥21
or,
(a−b−c)2+(b−a−c)2+(c−a−b)2≥a2+b2+c2.(1)
Multiplying this out the left-hand side of the last inequality gives 3(a2+b2+c2)−2(ab+bc+ca). Therefore the inequality (1) is equivalent to 2[a2+b2+c2−(ab+bc+ca)]≥0, which is evident because
2[a2+b2+c2−(ab+bc+ca)]=(a−b)2+(b−c)2+(c−a)2.
Equalities hold if (b+c)2=2(b2+c2) and (c+a)2=2(c2+a2), that is, a=b=c.
Third Solution: Given a function f of three variables, define the cyclic sum
cyc∑f(p,q,r)=f(p,q,r)+f(q,r,p)+f(r,p,q)
We first convert the inequality into
2a2+(b+c)22a(a+2b+2c)+2b2+(c+a)22b(b+2c+2a)+2c2+(a+b)22c(c+2a+2b)≤5
Splitting the 5 among the three terms yields the equivalent form
cyc ∑3[2a2+(b+c)2]4a2−12a(b+c)+5(b+c)2≥0(2)
The numerator of the term shown factors as (2a−x)(2a−5x), where x=b+c. We will show that
3(2a2+x2)(2a−x)(2a−5x)≥−3(a+x)4(2a−x)(3)
Indeed, (3) is equivalent to
(2a−x)[(2a−5x)(a+x)+4(2a2+x2)]≥0
which reduces to
(2a−x)(10a2−3ax−x2)=(2a−x)2(5a+x)≥0
evident. We proved that
3[2a2+(b+c)2]4a2−12a(b+c)+5(b+c)2≥−3(a+b+c)4(2a−b−c)
hence (2) follows. Equality holds if and only if 2a=b+c,2b=c+a,2c=a+b, i.e., when a=b=c.
Fourth Solution: Given a function f of three variables, we define the symmetric sum
sym∑f(x1,…,xn)=σ∑f(xσ(1),…,xσ(n))
where σ runs over all permutations of 1,…,n (for a total of n ! terms). For example, if n=3, and we write x,y,z for x1,x2,x3,
sym ∑x3sym ∑x2ysym ∑xyz=2x3+2y3+2z3=x2y+y2z+z2x+x2z+y2x+z2y=6xyz
We combine the terms in the desired inequality over a common denominator and use symmetric sum notation to simplify the algebra. The numerator of the difference between the two sides is
sym ∑8a6+8a5b+2a4b2+10a4bc+10a3b3−52a3b2c+14a2b2c2
Recalling Schur's Inequality, we have
=a3+b3+c3+3abc−(a2b+b2c+ca+ab2+bc2+ca2)a(a−b)(a−c)+b(b−a)(b−c)+c(c−a)(c−b)≥0
or
sym ∑a3−2a2b+abc≥0
Hence,
0≤14abcsym ∑a3−2a2b+abc=14sym ∑a4bc−28a3b2c+14a2b2c2
and by repeated AM-GM Inequality,
0≤sym ∑4a6−4a4bc
(because a46+a6+a6+a6+b6+c6≥6a4bc and its analogous forms)
and
0≤sym ∑4a6+8a5b+2a4b2+10a3b3−24a3b2c
Adding these three inequalities yields the desired result.
The problems on this page are the property of the MAA's American Mathematics Competitions