An unexpectedly loose-looking bound that makes the IMO inequality collapse

YoshiStoleMyTax·May 23, 2025(edited)

Bounding each summand by a constant loses the relationship

x^n+y^n=1

. Has anyone found a termwise estimate that actually uses that constraint? Problem: Let

n

be a positive integer, and let

x

and

y

be positive real numbers such that

x^{n}+y^{n}=1

. Prove that

\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{1}{(1-x)(1-y)} .

Source: IMO 2007 Shortlist · MNET-06P8

Open MNET-06P8 in the question bank →

4 replies

pixelated_toast·May 23, 2025(edited)

Try

(1+t^2)/(1+t^4)<1/t

for

0<t<1

. After multiplying positive denominators, the difference is

(1-t)(1-t^3)>0

YoshiStoleMyTax·May 23, 2025(edited)

That bound blows up when

t

is small. Wouldn't it make the sum much worse?

pixelated_toast·May 25, 2025(edited)

Individually yes, but

\sum_{k=1}^n x^{-k}=(1-x^n)/(x^n(1-x))=y^n/(x^n(1-x))

. The analogous y-sum has exactly the reciprocal ratio.

YoshiStoleMyTax·May 25, 2025(edited)

Multiplying cancels

x^n/y^n

and leaves

1/((1-x)(1-y))

. Both termwise inequalities are strict, so no equality case needs to be chased.

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An unexpectedly loose-looking bound that makes the IMO inequality collapse

YoshiStoleMyTax·May 23, 2025(edited)

Bounding each summand by a constant loses the relationship

x^n+y^n=1

. Has anyone found a termwise estimate that actually uses that constraint? Problem: Let

n

be a positive integer, and let

x

and

y

be positive real numbers such that

x^{n}+y^{n}=1

. Prove that

\left(\sum_{k=1}^{n} \frac{1+x^{2 k}}{1+x^{4 k}}\right)\left(\sum_{k=1}^{n} \frac{1+y^{2 k}}{1+y^{4 k}}\right)<\frac{1}{(1-x)(1-y)} .

Source: IMO 2007 Shortlist · MNET-06P8

Open MNET-06P8 in the question bank →

4 replies

pixelated_toast·May 23, 2025(edited)

Try

(1+t^2)/(1+t^4)<1/t

for

0<t<1

. After multiplying positive denominators, the difference is

(1-t)(1-t^3)>0

YoshiStoleMyTax·May 23, 2025(edited)

That bound blows up when

t

is small. Wouldn't it make the sum much worse?

pixelated_toast·May 25, 2025(edited)

Individually yes, but

\sum_{k=1}^n x^{-k}=(1-x^n)/(x^n(1-x))=y^n/(x^n(1-x))

. The analogous y-sum has exactly the reciprocal ratio.

YoshiStoleMyTax·May 25, 2025(edited)

Multiplying cancels

x^n/y^n

and leaves

1/((1-x)(1-y))

. Both termwise inequalities are strict, so no equality case needs to be chased.