Not simple

Algebra Level 4

If x + y + z ≤ 1 x+y+z\leq 1 , x , y , z > 0 x,y,z> 0

Find minimum of x 2 + 1 x 2 + y 2 + 1 y 2 + z 2 + 1 z 2 \sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}} Note:If the answer is 1.111111...Type 1.11


The answer is 9.05.

Son Nguyen by Son Nguyen , 20, Vietnam

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3 solutions

Otto Bretscher
Nov 12, 2015

If we let f ( x ) = x 2 + 1 x 2 f(x)=\sqrt{x^2+\frac{1}{x^2}} , then f ( x ) + f ( y ) + f ( z ) ≥ 3 f ( x + y + z 3 ) ≥ 3 f ( 1 / 3 ) ≈ 9.055 f(x)+f(y)+f(z)\geq 3f\left(\frac{x+y+z}{3}\right)\geq 3f(1/3)\approx \boxed{9.055} by Jensen's Inequality

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Don't you think that Jensen's is a contrived solution like AMGM?

Haha JK, good solution as usual. Comrade Otto

Pi Han Goh - 5 years, 6 months ago

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These inequalities are not contrived in and of themselves, but the way they are sometimes used is very contrived indeed. Here, Jensen jumps right at you. (Besides, AM-GM is a special case of Jensen, of course.)

I like Jensen's inequality because I can SEE it happen on the graph, in terms of concavity.

As an educator, I hate to see a lot of talented young people waste a lot of their time and energy on these silly inequalities rather than focusing on the important stuff: Calculus, linear algebra, etc.

Otto Bretscher - 5 years, 6 months ago

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Hmmmm... I wonder how many inequalities you consider to be contrived here .

I'm not going to lie and say that I enjoy these inequalities. As a matter of fact, I really dislike them (even Jensen's). This is a just a preference because I can't "savor the moment". But I still can see the use and beauty of them. Can't you?

As an educator, I hate to see a lot of talented young people waste a lot of their time and energy on these silly inequalities

Are you implying that all these inequalities are completely useless? And that they are esoteric and have no real application? If yes, what other fields you consider to be useless? (I'm curious)

Pi Han Goh - 5 years, 6 months ago

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@Pi Han Goh – Let me put it this way: The weight that certain topics (classic inequalities, Fermat's little theorem, and Newton Sums, for example) are given at Brilliant and in some math competitions is vastly different from the weight they have in mainstream higher education and in research math. In 40 years of enjoying math (as a student, teacher, and author), I have never seen anybody use "AM-GM" or "Newton Sums", for example, and Fermat only a few times.

Frankly, I don't see any beauty in contrived and "forced" math, although there may be some technical cleverness on the part of the setter and the solver of the problem. For me, mathematical beauty manifests itself in simplicity, the use of unifying concepts, surprising connections to other problems, and the use of symmetry, for example.

Otto Bretscher - 5 years, 6 months ago

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@Otto Bretscher – Maybe mainstream higher education should follow us haahahah!

Your comment resembles what Patrick Corn said in his reply to my comment . Thanks for the dialogue!

Pi Han Goh - 5 years, 6 months ago

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@Pi Han Goh – I'm happy to see that Patrick Corn has similar views on the subject... he is a wise man and his Erdos number is lower than mine ;)

Like Patrick, I was strongly influenced by the thinking of Comrade Alexander Grothendieck, who was my academic grand father (advisor of my advisor, Pierre Gabriel). Sadly, there has been a sharp decline in mathematical sophistication ;)

Otto Bretscher - 5 years, 6 months ago
Son Nguyen
Jan 12, 2016

By applying result of C-S inequality: x 2 + 1 x 2 + y 2 + 1 y 2 + z 2 + 1 z 2 ≥ ( x + y + z ) 2 + ( 1 x + 1 y + 1 z ) 2 \sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\geq \sqrt{(x+y+z)^{2}+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}} Since ( x + y + z ) ( 1 x + 1 y + 1 z ) ≥ 9 (x+y+z)\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )\geq 9 x + y + z ≤ 1 ⇒ 1 x + 1 y + 1 z ≥ 9 x+y+z\leq 1\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 9 And ( x + y + z ) 2 + 1 81 ( 1 x + 1 y + 1 z ) 2 ≥ 2 9 ( x + y + z ) ( 1 x + 1 y + 1 z ) = 2 (x+y+z)^2+\frac{1}{81}\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )^{2}\geq \frac{2}{9}(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=2 So ( x + y + z ) 2 + 1 81 ( 1 x + 1 y + 1 z ) 2 + 80 81 ( 1 x + 1 y + 1 z ) 2 ≥ 2 + 80 81 × 81 = 82 (x+y+z)^2+\frac{1}{81}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}+\frac{80}{81}\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )^{2}\geq 2+\frac{80}{81}\times 81=82 Hence ( x + y + z ) 2 + ( 1 x + 1 y + 1 z ) 2 ≥ 82 ≈ 9 , 05 \sqrt{(x+y+z)^2+(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}\geq \sqrt{82}\approx 9,05 The inequality hold when x = y = z = 1 3 x=y=z=\frac{1}{3}

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Ramiel To-ong
Dec 8, 2015

nice problem

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