Inspired by Ilham Saiful

Algebra Level 4

If a , b a,b are distinct positive real numbers, what is the solution set to

1 1 + a 2 + 1 1 + b 2 ≤ 2 1 + a b ? \frac{1}{ 1+a^2 } + \frac{1}{ 1 + b^2 } \leq \frac{ 2} { 1 + ab } ?

Notes:

  • The domain of the problem is also an unwritten constraint on the solution set. IE we're only interested in distinct, positive real numbers.

  • Yes, I'm aware that the inequality holds for all pairs a = b a = b , but this isn't in our domain.

Inspiration .

a , b ≤ 1 a, b \leq 1 a b ≤ 1 ab \leq 1 a 2 + b 2 ≤ 2 a^2 + b^2 \leq 2 All possible values None of the rest

Calvin Lin by Calvin Lin

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1 solution

Tom Engelsman
Sep 1, 2018

Let us re-write the above inequality as:

1 1 + a 2 + 1 1 + b 2 ≤ 2 1 + a b ; \frac{1}{1+a^2} + \frac{1}{1+b^2} \le \frac{2}{1+ab};

or ( 1 + b 2 ) + ( 1 + a 2 ) ( 1 + a 2 ) ( 1 + b 2 ) ≤ 2 1 + a b ; \frac{(1+b^2) + (1+a^2)}{(1+a^2)(1+b^2)} \le \frac{2}{1+ab};

or ( 2 + a 2 + b 2 ) ( 1 + a b ) ≤ 2 ( 1 + a 2 ) ( 1 + b 2 ) ; (2 + a^2 + b^2)(1+ab) \le 2(1+a^2)(1+b^2);

or 2 + a 2 + b 2 + 2 a b + a 3 b + a b 3 ≤ 2 + 2 a 2 + 2 b 2 + 2 a 2 b 2 ; 2 + a^2 + b^2 + 2ab + a^3b + ab^3 \le 2 + 2a^2 + 2b^2 + 2a^2b^2;

or 0 ≤ ( a 2 − 2 a b + b 2 ) − a b ( a 2 − 2 a b + b 2 ) ; 0 \le (a^2 - 2ab + b^2) - ab(a^2 - 2ab + b^2);

or 0 ≤ ( 1 − a b ) ( a 2 − 2 a b + b 2 ) ; 0 \le (1-ab)(a^2 - 2ab + b^2);

or 0 ≤ ( 1 − a b ) ( a − b ) 2 . 0 \le ( 1-ab)(a-b)^2.

Clearly, ( a − b ) 2 > 0 (a-b)^2 > 0 for all a , b ∈ R + , a ≠ b . a, b \in \mathbb{R^{+}}, a \neq b. This in turn forces 1 − a b ≥ 0 ⇒ a b ≤ 1 . 1 - ab \ge 0 \Rightarrow \boxed{ab \le 1}.

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That's the approach I took. I was trying to find a "direct classical inequality application" to this problem given how "nice" it looks.

That it is solely dependent on that value of a b ab was quite surprising to me.

Calvin Lin Staff - 2 years, 9 months ago

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Yup, it's straightforward and still gets the job done, Calvin.....thanks for the comment! I also updated the solution to reflect the domain of distinct, positive reals a and b too.

tom engelsman - 2 years, 9 months ago

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