A fourth-degree-looking inequality after switching to sum and difference

zipline_zebra·Dec 2, 2025(edited)

Expanding powers of a and b makes this uglier. Is the substitution

(a+b)/2,(a-b)/2

useful here, or does it only disguise the same algebra? Problem: Consider four positive real numbers

a

b

c

and

d

, satisfying

a^2 + ab + b^2 = 3c^2 \quad \text{and} \quad a^3 + a^2b + ab^2 + b^3 = 4d^3.

Prove that

a + b + d \le 3c.

Source: Baltic Way 2015 Shortlisted Problems · MNET-01CC

2 replies

shieldonbreak·Dec 2, 2025(edited)

Let x=(a+b)/2 and y=(a−b)/2. Then

c^2=x^2+y^2/3

and

d^3=x(x^2+y^2)

. The target becomes

2x+d\le3c

. You now know c≥x>0 and d≥x.

zipline_zebra·Dec 2, 2025(edited)

I get

(3c-2x)^2=3c^2-2x^2+6(c-x)^2\ge x^2+y^2

. Since

d^3=x(x^2+y^2)

and d≥x, we have

d^2\le x^2+y^2

. Both sides of the desired square-root comparison are positive, so

3c-2x\ge d

. That finishes it.

A hint, a different approach, or a thoughtful question can help.

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A fourth-degree-looking inequality after switching to sum and difference

zipline_zebra·Dec 2, 2025(edited)

Expanding powers of a and b makes this uglier. Is the substitution

(a+b)/2,(a-b)/2

useful here, or does it only disguise the same algebra? Problem: Consider four positive real numbers

a

b

c

and

d

, satisfying

a^2 + ab + b^2 = 3c^2 \quad \text{and} \quad a^3 + a^2b + ab^2 + b^3 = 4d^3.

Prove that

a + b + d \le 3c.

Source: Baltic Way 2015 Shortlisted Problems · MNET-01CC

Open MNET-01CC in the question bank →

2 replies

shieldonbreak·Dec 2, 2025(edited)

Let x=(a+b)/2 and y=(a−b)/2. Then

c^2=x^2+y^2/3

and

d^3=x(x^2+y^2)

. The target becomes

2x+d\le3c

. You now know c≥x>0 and d≥x.

zipline_zebra·Dec 2, 2025(edited)

I get

(3c-2x)^2=3c^2-2x^2+6(c-x)^2\ge x^2+y^2

. Since

d^3=x(x^2+y^2)

and d≥x, we have

d^2\le x^2+y^2

. Both sides of the desired square-root comparison are positive, so

3c-2x\ge d

. That finishes it.