let $a,b\in R$,and such that $$\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$$

prove that $$a+b\ge 0$$

I think this is very beatifull problem, have you nice methods? Thank you,

I have see this problem $$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$$ then we have $$x+y=0$$

This problem have some nice methods, let $f(x)=\ln\left({x+\sqrt{x^2+1}}\right)$, then $f(x)$ is increasing and is odd funciotion, and $$\ln\left({y+\sqrt{y^2+1}}\right)=-\ln{-y+\sqrt{y^2+1}}=-f(-y)$$ so $$f(x)+f(-y)=0\Longrightarrow f(x)=f(-y)$$ then $x+y=0$

and I have see this problem $$\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1$$ then we have $$x+y=0$$ solution: let $$x=\dfrac{1}{2}\left(u-\dfrac{1}{u}\right),y=\dfrac{1}{2}\left(v-\dfrac{1}{v}\right)$$

then we have $$\dfrac{(uv-1)((u+v)^2uv+(u-v)^2)}{u^2v^2}=0$$

since $$(u+v)^2uv+(u-v)^2\ge 0$$ so $$uv=1$$ then $$x+y=\dfrac{1}{2}\left(u+v-\dfrac{1}{u}-\dfrac{1}{v}\right)=0$$

  • $\begingroup$ I believe you want brackets $f(x) = \ln ( \ldots )$. $\endgroup$
    – Calvin Lin
    May 15, 2013 at 6:24
  • $\begingroup$ Did you try some substitutions like $x=ab, y =\frac a b $? $\endgroup$
    – gukoff
    May 15, 2013 at 6:28
  • $\begingroup$ @math110 I've added on how to approach the equality case using my solution, which is more motivated than your $f(x)$ approach. $\endgroup$
    – Calvin Lin
    May 15, 2013 at 6:39
  • $\begingroup$ @math110 You might want to see here for other approaches to your second problem. $\endgroup$
    – Calvin Lin
    May 27, 2013 at 20:33

1 Answer 1


Step 1: Multiply by the conjugate, we get that

$$ b^4 a^4 \leq a^2 b^2 ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b), $$ $$ a^2 b^2 \leq ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b)$$.

This gives us the chain of inequalities

$$ ( \sqrt{a^2 + b^4} -a ) ( \sqrt{ b^2 + a^4} - b) \leq a^2 b^2 \leq ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b)$$

Take the extreme ends, we get $0 \leq a \sqrt{b^2 + a^4} + b \sqrt{a^2 + b^4} $.

Step 2: Consider cases

If $a, b <0$, then the terms in the LHS are clearly greater than $b^2 \times a^2$, hence the inequality is not true.

If $a,b \geq 0$, then clearly $a+b \geq 0.$

Hence, we may assume that $ a \leq 0 \leq b$, and we want to show that $ -a \leq b$. Now, because we are used to dealing with positive (non-negative reals), let me replace $a$ with $-a$ (not necessary, but simplifies considerations later)

Step 3: With this substitution, the inequality in step 1 gives us

$$ a\sqrt{b^2 + a^4} \leq b \sqrt{a^2 + b^4}$$

Since the LHS is non-negative, we may square it to obtain

$$a^2 (b^2 + a^4) \leq b^2 (a^2 + b^4 \Rightarrow a^6 \leq b^6 \Rightarrow a \leq b.$$

But this is what we want to show in Step 2, hence we are done. (remember we substituted $a$ for $-a$.)

Step 1 gives you another way to prove your equality case. Namely, we get that

$$ 0 = y \sqrt{x^2+1} + x \sqrt{y^2+1} $$

Hence, we have $ - y \sqrt{x^2 + 1} = x \sqrt{y^2+1} \Rightarrow y^2(x^2+1)=x^2(y^2+1) \Rightarrow y^2=x^2$. Then check that $y=x$ is not a valid solution (unless $y=x=0$), hence we must have $y=-x$.

This seems much more direct than your approach, and is motivated by considering conjugates.

The inequality in step 1 can also be obtained directly, by expanding and showing that

$$ 0 \leq \sqrt{a^2+b^4} \sqrt{b^2+a^4} + ab - a^2b^2 \leq a \sqrt{b^2 + a^4} + b \sqrt{a^2 + b^4}$$

However, this is not immediately obvious from the question. It's more of 'on hindsight'.

I'd be interested in seeing how the equality case can be obtained through direct expansion. (I don't see how to do this as yet.)


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