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Let $a, b \in \mathbb{R}$. Show that $|ab| \le \frac 12 (a^2 + b^2 )$.

Need consider three cases:
(i) both $a,b \ge 0$: $|ab| = ab=\frac{2ab}2= \frac{(a+b)^2-a^2-b^2}2=\frac{(a+b)^2-(a^2+b^2)}2$.

(ii) both $a,b \lt 0$: $|ab| = ab$ , same as in (i).

(iii) either one of $a,b \lt 0$: $|ab| = -ab=-\frac{2ab}2= -(\frac{(a+b)^2-a^2-b^2}2)=-(\frac{(a+b)^2-(a^2+b^2)}2)$.


So, need prove $2ab \le a^2+b^2$ which is again back to the question.

I have in mind only an approach based on triangle formed by three vectors, but am unclear about using that. The reason is need to constrain the angle formed to $90^o$, as only then $a^2 + b^2 = a^2+b^2$.
But, even with that constraint, the issue is that the vector approach has equivalent formulation only in terms of complex quantities.

So, no progress possible.

jiten
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    If the reason for closing is it being 'duplicate', then it should apply 'also' to the post causing my post's closure; as there also it is declared as duplicate. Else, the reason should be made clear. – jiten Jun 01 '19 at 08:26
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    Indeed, it should. A big problem I see here that even though you posted this as a proof-verification question, there is A) no proof to be verified, and B) the answerers treated it as a duplicate, and reproduced the same proof as in those old threads. To a great extent your answerers prove that this is a duplicate in spite of your attempt. – Jyrki Lahtonen Jun 01 '19 at 08:27
  • Judging from the positive reaction to my recent meta post it is safe to conclude that I am not alone seeing the problem this way. I am mostly pointing my finger at the answerers. As veterans of the site they have no excuse for not knowing that this common inequality must have been handled many times on our site. – Jyrki Lahtonen Jun 01 '19 at 08:29
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    Anyway, we have more than 10 threads dedicated to essentially the same inequality. If a new users recognized it as something they can prove, and jumps to jot a proof down, I understand why that may happen. But a user with more than 20k rep should know better. – Jyrki Lahtonen Jun 01 '19 at 08:32
  • @JyrkiLahtonen I tried to write post's title (as the new post composing process shows then the related posts) in umpteen number of ways, apart from searching with query in google specifically for mse. But, did not get any of the duplicates. I have removed the pointed out tag. The reason the tag was put is that there is nothing related to proof that fitted better. – jiten Jun 01 '19 at 08:40
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    It would have been fine, if only the answerers studied your suggested line of looking at triangles. But, no, they wanted to grab a low-hanging fruit. Actually the law of cosines would allow you to complete that approach. If the claimed inequality were not true, then selecting a suitable angle between the sides of lengths $a$ and $b$ would lead to the absurd situation that the squared length of $c$ would be negative. Let's wait for the other voters to tell their opinion. I would be fine with someone posting an answer along those lines, but getting the current answers deleted. – Jyrki Lahtonen Jun 01 '19 at 08:46
  • In other words, the question became a duplicate when you accepted an answer. – Jyrki Lahtonen Jun 01 '19 at 08:52
  • @JyrkiLahtonen Have a point - my post was never a copy, nor aware of. Also, my approach using vectors is not shown in the two posts at : 1) https://math.stackexchange.com/q/470221/424260, 2) https://math.stackexchange.com/q/320244/424260. Even the selected answer is different from any other answer on these posts. – jiten Jun 01 '19 at 09:45
  • @JyrkiLahtonen Please elaborate your answer using law of cosines. I am not clear about that. I hope you mean that only one angle is possible, but very confused about that. I also wanted to create response using vectors on that lines, but my post is wrong in stating that only vector representation is possible, that necessitates a complex number representation. I mean that scalars (magnitudes along $x,y$ axis) can also be used and then cosine law can be applied, without its equivalent rep. in complex plane. – jiten Jun 01 '19 at 09:52
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    If $2|ab|>a^2+b^2$ then there exists an angle $\theta, 0<\theta<\pi$ such that $$a^2+b^2-2ab\cos\theta<0.$$ Gotta rush, sorry. – Jyrki Lahtonen Jun 01 '19 at 11:17

2 Answers2

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As you've already noted, we can restate the claim as $a,\,b\ge 0\implies 2ab\le a^2+b^2$. This follows from $a^2+b^2-2ab=(a-b)^2\ge0$.

J.G.
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We focus on $a, b > 0$.

Suppose on the contrary that we have

$$2ab > a^2 + b^2$$

but from the cosine rule, we have $$a^2 + b^2 - 2ab\cos \theta \ge 0$$

In particular, let $\cos \theta = 1$, then

$$a^2+b^2 \ge 2ab$$

which is a contradiction.

Old answer:

We have $$(|a|-|b|)^2 \ge 0$$

$$|a|^2 - 2|a||b| + |b|^2 \ge 0$$

$$a^2 - 2|a||b| + b^2 \ge 0$$

$$2|a||b| \le a^2 + b^2$$

Dividing by $2$,

$$|a||b| \le \frac12 (a^2 + b^2)$$

Siong Thye Goh
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  • Have another smallish question like this, from the same source: Let $a, b \in \mathbb{R}$. Show that $| |a| − |b| | \le |a − b|$. I hope here need consider all four cases: (i) $a\ge 0, b\ge 0$, (ii) $a\ge 0, b\lt 0$, (iii) $a\lt 0, b\ge 0$, (i) $a\lt 0, b\lt 0$. For (i) $|a-b|=|a-b|$, (ii) $|a+b|\lt |a-b|$, & so on. – jiten Jun 01 '19 at 07:43
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    $|a| =|a-b+b|\le |a-b|+ |b|$, so we have $|a|- |b| \le |a-b|$ also, you can swith the role of $a$ and $b$ and conclude that. Nothing wrong with consider cases but it's good to have a shorter proof sometimes. This is called the reverse triangle inequality. – Siong Thye Goh Jun 01 '19 at 07:48
  • btw, let me make life easier. just unaccept this solution and I will delete the answer. – Siong Thye Goh Jun 01 '19 at 09:16
  • I amjust leaving the answer here for $15$ minutes. To be deleted. – Siong Thye Goh Jun 01 '19 at 09:58
  • Please do not delete. This is a good & unique answer. It also helped me understand the subtlety of the comments given. Might be needs some detail about stating that my approach was wrong as vectors are not a must, scalars can also be used. Then scalars (magnitudes) are enough, rather than vectors (complex plane rep.). – jiten Jun 01 '19 at 10:03
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    regardless of whether you viewed as vector or complex number, I just used a result that uses the length. – Siong Thye Goh Jun 01 '19 at 10:12
  • Request another two questions (with actual question no. from the same source): Q.2. Let $f(x) = \frac{7-\sqrt{x^2-9}}{\sqrt{25-x^2}}$. Then $dom f = (-5,-3]\cup[3,5)$. There is open interval for denominator not to become $0$.

    Q.3: Find the domain & range of the function $f(x)= 2\sqrt{4-x^2} -3$
    $dom f = [-2,2]; ran f =[-3,1] $, with range $=1$ at $x=0$, while range $=-3$ at $x=-2,2$. I have not considered the graph of the function, but the range values show that it is neither increasing nor decreasing in the given domain. But, is there any other way to know range.

     – jiten Jun 02 '19 at 04:43
  • Let us continue this discussion in chat. – Siong Thye Goh Jun 02 '19 at 09:44
  • Need help in desmos to draw function $a = ln|x^3+1|$, or if possible in a P.L. as python. Can tell in chat too. I am worried as to see the simple integral (with result given above for graphing as $'a'$) properties, need processor & memory intensive s/w as matlab, octave, etc. I feel it (graphical rep.) is needed to better grasp integrals. But, it is difficult to dedicate one machine for octave/ matlab/ any other heavyweight s/w. – jiten Jun 04 '19 at 03:04
  • Have made at https://www.desmos.com/calculator/jf8hglwe0q, but not clear how to see it as integral of $\frac{3x^2}{x^3+1}dx$. Definitely need chat, as visualiing the integral in the graphical form has a total disconnect. – jiten Jun 04 '19 at 06:12
  • Made graph for integration of $\frac{3x^2}{x^3+1}$ at https://www.desmos.com/calculator/ijxrdqegd9. It is modifn. of some existing one at desmos (https://www.desmos.com/calculator/cprmiyx4i9). I see no similarity even with that of $a=ln|x^3+1|$. – jiten Jun 04 '19 at 07:20
  • Please help to find solution for $x = \cot(\frac 1x)$, for $x$ in the interval $0\le x \le \frac 1{\pi}$. Had to resort to first 3 terms of series of $\cot(z)= \frac 1z - \frac{z}3 - \frac{z^3}{45} \implies z = z - \frac 1{(z)(3)} - \frac 1{(z^3)(45)} \implies z^2 = -\frac 1{15}$, by substituting for $z$. Taking $z = x+iy$, get : $x^2-y^2 = -\frac 1{15}, xy =0$. Assuming $y=0$, get imaginary (non-real) values for $x$. Assuming $x=0$, get $y = \sqrt{\frac 1{15}}$. Unable to solve further. – jiten Jun 05 '19 at 07:01
  • Please help with the above comment. If the series expansion is wrong for $x = cot(1/x)$, as need $z$ rather than $x$, then please tell. I mean that using complex number based formula for $x$ is correct or not. – jiten Jun 05 '19 at 08:14
  • Have posted at : https://math.stackexchange.com/q/3251623/424260. Please help with it. – jiten Jun 05 '19 at 08:37
  • can I have a small chat? – jiten Jun 05 '19 at 17:35