Advanced book on Mathematics Olympiad

Algebra 381

E(a+α, b, c−α)=E(a, b, c)+α( 1 − 2 b)[(c−a)−α].

Sinceb≤canda+b+c=1, we haveb≤^12. This means thatE(a+α, b, c−α)≥
E(a, b, c). So we were able to make one ofaandcequal to^13 by increasing the value
of the expression. Repeating the argument for the remaining two numbers, we are able
to increaseE(a, b, c)toE(^13 ,^13 ,^13 )= 277. This proves the inequality.
(communicated by V. Grover)

134.The inequality from the statement can be rewritten as

∏n

∏ j=^1 xj

n

j= 1 (^1 −xj)

≤

(∑

n

j= 1 xj

)n

(∑

n

j= 1 (^1 −xj)

)n.

If we fix the sumS=x 1 +x 2 +···+xn, then the right-hand side is constant, being equal
to(n−SS)n. We apply Sturm’s principle to the left-hand side. If thexj’s are not all equal,
then there exist two of them,xkandxl, withxk<Sn <xl. We would like to show that
by adding a small positive numberαtoxkand subtracting the same number fromxlthe
expression grows. This reduces to

(xk+α)(xl−α)

( 1 −xk−α)( 1 −xl+α)

<

xkxl

( 1 −xk)( 1 −xl)

.

Some computations transform this into

α( 1 −xk−xl)(xl−xk−α) > 0 ,

which is true ifα<xl−xk. Choosingα=xl−Snallows us to transformxlintoSnby
this procedure. One by one we make the numbers equal toSn, increasing the value of the
expression on the left each time. The fact that in this case we achieve equality proves the
inequality in the general case.
(Indian Team Selection Test for the International Mathematical Olympiad, 2004)

135.We apply the same kind of reasoning, varying the parameters until we reach the
maximum. To find the maximum of

√

a+

√

b+

√

c+

√

d, we increase the suma+b+c+d
until it reaches the upper limit 30. Becausea+b+c≤14 it follows thatd≥16. Now
we fixa, band varyc, dto maximize

√

c+

√

d. This latter expression is maximal if
canddare closest toc+ 2 d. But sincec+d≤30,c+ 2 d≤15. So in order to maximize
√
c+

√

d, we must choosed=16.
Now we havea+b+c=14,a+b≤5, anda≤1. The same argument carries
over to show that in order to maximize

√

a+

√

b+

√

cwe have to choosec=9. And
the reasoning continues to show thatahas to be chosen 1 andbhas to be 4.
We conclude that under the constraintsa ≤1,a+b ≤5,a+b+c≤14, and
a+b+c+d ≤30, the sum

√

a+

√

b+

√

c+

√

dis maximal whena=1,b=4,