We will prove that the claim follows from these two inequalities √ 1 1 + x + 1 √1 + y ≥ 2 p1 + √xy for x, y &gt

Problem 10944

(American Mathematical Monthly, Vol.109, May 2002) Proposed by M. Mazur (USA).

Prove that ifa, b, c are positive real numbers such that abc ≥ 2⁹, then

√ 1

1 + a+ 1

√1 + b + 1

√1 + c ≥ 3 p1 +√³

abc

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We will prove that the claim follows from these two inequalities

√ 1

1 + x + 1

√1 + y ≥ 2

p1 + √xy for x, y > 0 and xy ≥ 9, (1)

√ 2

1 + x+ 1

√1 + y ≥ 3 q

1 +px^{3 2}y

for x, y > 0 and x²y ≥ 2⁹. (2)

In fact, let a ≥ b ≥ c > 0 be such that abc ≥ 2⁹then ab ≥ 9 (otherwise abc < 27) and we can apply (1) to the first two terms. Afterwards we use (2) with x =√

ab and y = c completing the proof

√ 1

1 + a+ 1

√1 + b+ 1

√1 + c ≥ 2 p1 +√

ab+ 1

√1 + c ≥ 3 p1 +√³

abc.

Now we prove that the inequalities (1) and (2) hold.

1) Let’s consider the function

f (x) = 1

√1 + x + 1

p1 + p/x for x > 0 and p ≥ 9 then it suffices to show that

x>0inff (x) = f (√

p) = 2

p1 + √p. The derivative of f is

f^′(x) = −1

2· (1 + x)^−3/2−1

2· (1 + p x)^−3/2·



−p x²



which is non-negative for x > 0 iff

p²(1 + x)³− x(x + p)³= (p − x²)(x²− p(p − 3)x + p) ≥ 0.

The condition p ≥ 9 forces the second quadratic polynomials to have two distinct positive roots.

Since their product is equal to p then the smaller root belongs to the interval (0, √p) and the other one stays in (√p, +∞). So the sign of f^′ is easily determined and it follows that, for x > 0, f has a unique local minimum at √p. After checking the boundary behaviour of f , we can conclude that

√p is just the global minimum

x→0+lim f (x) = lim

x→+∞f (x) = 1 ≥ f(√p) = 2

p1 + √p because p ≥ 9.

2) Now we consider the function

g(x) = 2

√1 + x + 1

p1 + q/x² for x > 0 and q ≥ 2⁹ then it suffices to show that

x>0inf g(x) = g(√³q) = 3 p1 +√q³ . Again we compute its derivative

g^′(x) = −(1 + x)^−3/2− (1 + q x²)^−3/2·



−q x³



which is non-negative for x > 0 iff

q^2/3(1 + x) − (x²+ q) = −x²+ q^2/3x − (q − q^2/3) ≥ 0.

Since q ≥ 2⁹ then this quadratic polynomial has two distinct positive roots: the smaller one is √q³ and the other one is √q(³ √q −1). So the sign of g^{3 ′}is easily determined and it follows that, for x > 0, g has a unique local minimum at√q. After checking the boundary behaviour of g, we can conclude³ that √q is just the global minimum³

x→0+lim g(x) = 2 > lim

x→+∞g(x) = 1 ≥ g(√³q) = 3

p1 +√q³ because q ≥ 2⁹.