Going through solutions of IMO'09. Bumped into a without-loss-of-generality assumption that I can't comprehend.  Here's the statement of the problem:  Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that $$ \frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}\leq\frac{3}{16}. $$  Here's how they start in the solution:  We prove the homogenized inequality $$ \frac{(a+b+c)^2}{(2a+b+c)^2}+\frac{(a+b+c)^2}{(2b+c+a)^2}+\frac{(a+b+c)^2}{(2c+a+b)^2}\leq\frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $$ for all positive real numbers $a,b,c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a,b,c&gt;0$, fulfilling this condition, the inequality $$ \frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\leq\frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). $$ And so on... If someone's interested in the rest, it's problem A2, solution 2 in this pdf:   Why is there no loss of generality in such a choice of $a,b,c$?
122098
Refer to the problem below (IMO 2009 Shortlist)     Let $a, b, c$ be positive real numbers such that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = a + b+ c$$   Prove that    $$\frac{1}{(2a+b+c)^2} + \frac{1}{(2a+b+c)^2} + \frac{1}{(2a+b+c)^2} \leqslant \frac{3}{16} $$    In the solution given by the  (Pg 16), it states that      Without loss of generality, we choose $$a +b+c = 1$$.    Thus, the problem becomes    $$ \frac{1}{(1 + a)^2} + \frac{1}{(1 + b)^2} +\frac{1}{(1 + b)^2} \leqslant \frac{3}{16}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$$    Applying Jensen’s inequality to the function $$f(x) = \frac{x}{ (1+ x)^2}$$   ,which is concave for $0 ≤ x ≤ 2$   and increasing for $0 ≤ x ≤ 1$, we obtain   $$α \frac{a}{(1 + a)^2} + β \frac{b}{(1 + b)^2} + γ \frac{c}{(1 + c)^2} \leqslant (α + β + γ) \frac{A}{(1 + A)^2}$$   , where $A =\frac{αa + βb + γc}{α + β + γ}.$   Choosing $α = \frac{1}{a} , β =\frac{1}{b},$ and  $γ = \frac{1}{c}$   , we can apply the harmonic-arithmetic-mean inequality   $$A =\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} ≤ \frac{a + b + c }{3} = \frac{1}{3} &lt; 1$$   Finally we prove:   $$ \frac{1}{(1 + a)^2} + \frac{1}{(1 + b)^2} +\frac{1}{(1 + b)^2} \leqslant ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \frac{A}{(1 + A)^2} \leqslant ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c})\frac{A}{(1 + \frac{1}{3})^2} = \frac{3}{16}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $$   However,  (Pg 4) differs by having $a +b+c=3$ and setting $f(x) = \frac{1}{16x} - \frac{1}{(x+3)^2}$ and allowing Jensen to resolve the rest.   The questions are as follows;   How do you choose before hand which $a + b+ c =$ to choose?  Why particularly does the assumption differ between the 2 solutions? How does one know before hand the choosing of $α = \frac{1}{a} , β =\frac{1}{b},$ and  $γ = \frac{1}{c}$?  Where and how did $A =\frac{αa + βb + γc}{α + β + γ}$ come from?   Any help would be much appreciated.
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Hy all! I'm having trouble finding a proof for the following problem:  Show that $a^x+b^x+c^x&gt;(a+b+c)^x$, if $a,b,c&gt;0$ and $0&lt;x&lt;1$ (over the real numbers).  This inequality has been torturing me for long hours the least. I couldn't find anything on Google. This kind of equation isn't made for search engines tbh.  Anyway, I really thinked about it a lot but didn't make any significant progress. It would be great to see a proof without the really high-end Mathematics. Any ideas to start with?
65983
I'm trying to prove the following inequality by purely algebraic means  $$\frac{|x-y|}{1+|x-y|}\leq \frac{|x-z|}{1+|x-z|}+\frac{|z-y|}{1+|z-y|}.$$  My arguement is as follows:   Notice that, $|x-z|+|z-y| \geq |x-y|$ and from this inequality we can write $1+|x-z|+|z-y| \geq 1+|x-z| $ and $1+|x-z|+|z-y| \geq 1+|z-y|$. Now,   $$\frac{|x-y|}{1+|x-y|} \leq \frac{|x-z|+|z-y|}{1+|x-z|+|z-y|}\leq \frac{|x-z|}{1+|x-z|}+\frac{|z-y|}{1+|z-y|}.$$  Is this correct?
36735
Here there is my proof (quite short and easy) of a rather straightforward result. The text of this question comes from a previous question of mine, where I ended up working on a wrong inequality. Here things are fixed. Still, I would like to know:  if it is sound, because absolute value always creates me some problem, and if there is a shorter (neater) way to get the result (maybe without using contradiction).    Proposition: $|d(x,y)-d(y,z)| \leq d(x,z)$. Proof: We proceed by cases.  Case 1: $d(x,y)-d(y,z) \geq 0$. Assume by contradiction that $d(x,y)-d(y,z) &gt; d(x,z)$. Hence, $d(x,y)&gt; d(y,z) + d(x,z)$, contradicting the triangle inequality.  Case 2: $d(x,y)-d(y,z) &lt; 0$. Assume by contradiction that $ - d(x,y) + d(y,z) &gt; d(x,z)$. Hence, $d(x,y) &lt; d(y,z) - d(x,z)$. By applying the triangle inequality, we have that $d(y,x) \geq d(y,z) + d(x,z)$, which is by symmetry equivalent to  $d(x,y) \geq d(y,z) + d(x,z)$, obtaining the desired contradiction. $\square$     As always, any feedback is welcome. Thank you for your time.
1567
On , one can see that   $$\|fg\|_r\leq \|f\|_p\|g\|_q,$$ when $$\frac{1}{r} = \frac{1}{p} + \frac{1}{q}, \ p,q\in(0,\infty].$$  I have two questions,   Is the statement true? If so, is it stated in some references? I need it for my thesis.
88710
I have just learnt about homogenization, however I don't really understand it and I can't really utilize it ( this is where I learnt about it). Could you please explain to me how to use it optimally? Moreover I found a question on IMO 2020, which I believe can be solved using homogenization. It goes as follows: Let the real numbers $a, b, c, d$ be such that $a\ge b\ge c \ge d &gt;0$ and $a+b+c+d=1$. Prove that: $(a+2b+3c+4d)a^ab^bc^cd^d&lt;1$ I attempted to solve it by saying that RHS is equal to $(a+b+c+d)^a$, but then I realized that this doesn't help and it doesn't homogenize the inequality either. Then I tried using AM-GM on LHS and getting something which can be homogenized, however I did not succeed there either. Could you please explain to me how to use homogenization optimally and how this question could be solved using it?
97993
I was asked to prove an inequality: For any $n$ positive numbers $\{a_i\}$ with $a_{1}a_{2}\cdots a_{n} = 1$ and $m \geq n-1$ be a non-negative integer, $a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}$  After several attempts, I think I had to use induction on both $m$ and $n$. In particular, I fix an $n$ and claim that the inequality holds for every $m$, and use induction on $n$.  The case where $n = 2$ could be done by rewriting the inequality as $a_1^2 + a_2^2 \geq a_1 + a_2$ and assume  that $a_1 \leq 1 \leq a_2$.   In the inductive step, given $a_1^k + a_2^k \geq a_1 + a_2$ , by writing $a_1^{k+1} + a_2^{k+1} = (a_1^{k} + a_2^{k})(a_1 + a_2) - a_1^{k}a_2^{k} (a_1 + a_2)$, the result follows.  However, this method seems to fail for $n&gt;2$, as the right-hand side of the inequality becomes more complicated. Should I try not to prove it by induction?
91462
Prove that the general arithmetic-geometric mean inequality \begin{equation*} (a_{1}a_{2}...a_{n})^\frac{1}{n}\leq\frac{a_{1}+a_{2}+...+a_{n}}{n} \end{equation*} holds for all $a_{i}$ positive real numbers.  I keep getting stuck half way. This is review material for me (which I feel like I should be getting easily, but that's not the case unfortunately).
71836
Let $a, b, c, d &gt; 0$. Prove that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$.   I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated.
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I have to apologize in advance that the question is missing some vital information (like, $a,b,c$ are positive? natural? rational?) I took it off some video off youtube.  In the video, an 11 year old was in the process of solving this question: prove $\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+c^3+abc}+\frac{1}{b^3+c^3+abc} \leq \frac{1}{abc}$.  I don't really see how we can solve it in a simple way without using differential multivariable calculus. And even with calculus - it isn't so simple and not very pleasant. If anyone knows the question in its full form and knows how to answer it simply, I will be very happy to see it.  In the video, the child said "wlog $abc=1$" and defined $x=a^3$, $y=b^3$, $z=c^3$ and so he had to prove that  $\frac{1}{x+y+1}+\frac{1}{x+z+1}+\frac{1}{y+z+1} \leq 1$  The $abc=1$ part seems to be not so general, but if that helps you solve the question in any simple way - have at it.
41505
Let $a$, $b$ and $c$ be positive numbers. Prove that:   $$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$   This inequality is strengthening of the following Vasile Cirtoaje's one, which he  created in 2005.     Let $a$, $b$ and $c$ be positive numbers. Prove that:   $$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$   My proof of this inequality you can see here:   But this way does not help for the starting inequality.  A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is  $0.0000158...$.  I tried also to use Cauchy-Schwarz, but without success.  Also, I think the BW (see here  I tryed!) does not help.
76501
For $a,b,c&gt;0$, prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c}.$$ (Source : Balkan Mathematical Olympiad 2005)  My work: From the given inequality, we can have, $$\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\ge \frac{4(a-b)^2}{a+b+c}$$ This can be re-written as, $$(a+b+c)\left(\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}\right)\ge 4(a-b)^2$$ Now, what am I supposed to do? I am doubtful whether the question is correct or not. Please help. EDIT: After checking quite a few cases I conclude that the inequality is actually working. So there might not be any mistakes at all.
93272
It is also given that $abc = 1$. I used AM-GM inequality to reach till $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} $ How to go further
11089
Prove that $\forall a,b,c,d\in \mathbb{R}\,$,  $(a^2-b^2)(c^2-d^2)\leq(ac-bd)^2$ and equalities hold if and only if $ad=bc$.    I need hints in dealing with the inequality. I think what I have with dealing with the case of equality is sufficient.   $\begin{align}(a^2-b^2)(c^2-d^2)\leq(ac-bd)^2&amp;\Longleftrightarrow(ac)^2-(ad)^2-(bc)^2+(bd)^2\leq(ac)^2-2abcd+(bd)^2\\ &amp;\Longleftrightarrow-(ad)^2-(bc)^2\leq-2abcd \\ \end{align}$  Suppose $ad\neq bc$, then $-(ad)^2-(bc)^2\leq-2abcd\Longrightarrow 2abcd\leq(ad)^2+(bc)^2,\,$ since $0&lt;(ad)^2$ and $0&lt;(bc)^2$ if $ad\neq bc$. Thus, the equality is only satisfied if and only if $ad=bc$, since this implies:  $$-(ad)^2-(bc)^2=-(ad)^2-(ad)^2=-2(ad)^2\leq-2abcd=-2ad\cdot ad=-2(ad)^2.$$  Hints to show inequality? Thanks.
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How one can prove the following.  Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:  $(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$  WLOG one can assume that $0\le a\le b\le c$.  It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le a\le b\le c$.
86737
So I have to prove $$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$  I rearranged it   $$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$  My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?  Thanks
93377
Please help me for prove this inequality: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) &gt; 0$$