I am trying to prove that if $a&gt;1$ $x,y&gt;0$, and there is $b$ such that $\frac{1}{a} + \frac{1}{b} = 1$ then xy $\leq \frac{x^a}{a} + \frac{y^b}{b}$  I have made several attempts, but I get to no answer. Is there any inequality that needs to be shown before?
46332
OK guys I have this problem:  For $x,y,p,q&gt;0$ and $ \frac {1} {p} + \frac {1}{q}=1 $ prove that $ xy \leq\frac{x^p}{p} + \frac{y^q}{q}$  It says I should use Jensen's inequality, but I can't figure out how to apply it in this case. Any ideas about the solution?
38566
Prove that $\forall x,y&gt;0, x^x+y^y \geq x^y + y^x$   A friend of mine told me none of the teachers in my school have succeeded in proving this seemingly simple inequality (it was asked at an oral exam last year). I tried it myself, but I made no significant progress.
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
2180
Show that $|x-y|\leq |x|+|y|$.   I know this probably applies the triangle inequality in some way, but I can't figure it out.
6481
I would appreciate if somebody could help me with the following problem  Let $|x|\leq1, |y|\leq 1$. Show that  $$ 0  \leq x^{2}+y^{2} -2x^{2} y^{2}+2xy  \sqrt{1-x^{2}}  \sqrt{1-y^{2}} \leq 1 $$
88959
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$   First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expression as below;  \begin{align} \frac{x^2}{1+x^4} - \frac{1}{2} &amp;\leq 0\\\\ \impliedby\frac{2x^2 - (1+x^4)}{2(1+x^4)} &amp;\leq 0\\\\ \impliedby -\frac{(x^4 - 2x^2 + 1)}{2(1+x^4)}&amp;\leq 0\\\\ \impliedby \frac{x^4 - 2x^2 + 1}{2(1+x^4)} &amp;\geq 0\\\\ \end{align}   We then make the observation that $x^4 - 2x^2 + 1 = (x^2 - 1)^2 \geq 0$ for all $x \in \Bbb R$ and that $1 + x^4 \geq 1$ for all $x \in \Bbb R$, and so the final inequality holds for all $x \in \Bbb R. \\\hspace{92pt}\square$  Is it okay to prove it in this way? I have no solutions so it would be nice if I could have some validation!
85884
Show that for $x,y,z &gt; 0$ the inequality is true: $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+x+y+z \geq \frac{(x+y)^2}{y+z}+\frac{(y+z)^2}{z+x}+\frac{(z+x)^2}{x+y}$  I have tried Holder, but i had no luck. Please give me a hint of how to start.
70751
Let $a$, $b$ and $c$ be positive real numbers with $abc=1$. Prove that   $$ \frac{a^{n+2}}{a^n+(n-1)b^n}+\frac{b^{n+2}}{b^n+(n-1)c^n}+\frac{c^{n+2}}{c^n+(n-1)a^n} \geq \frac{3}{n} $$   for each integer $n$.   I have used Cauchy-Schwarz inequalities and Jensen inequality. But I am stuck. I need some idea and advice on this problem. Induction would be cruel.
80322
Prove: $$ (a+b)^\frac{1}{n} \le a^\frac{1}{n} + b^\frac{1}{n}, \qquad \forall n \in \mathbb{N} $$ I have have tried using the triangle inequlity $ |a + b| \le |a| + |b| $,  without any success.
17095
let a,b be real  numbers and $\varepsilon&gt;0$. prove the above inequality. I dont know how to even start. please, I need your help.
45865
Are there any positive integers $x,y,z$ such that $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1?$$ Prove/Disprove.   I've plugged in random positive integers for $x,y,z$ and I have not been able to get the equation to equal $1$.
31355
Suppose that $a$, $b$ and $c$ are positive real numbers and that $a \leq b + c$. By cross multiplying or otherwise, show that  $$\frac{a}{1 + a} \leq \frac{b}{1 + b} + \frac{c}{1 + c}$$  Anyone able to get this to work out? I cross multiplied them but am unable to manipulate it into an expression that verifies the statement.
88269
Show that $\frac c {1+c} \le \frac a {1+a} + \frac b {1+b}$ , for $c \le a+b$ and $a,b,c \ge 0$  So need to show $\frac c {1+c} \le \frac {a+b+2ab} {1+a+b+ab}$  We have $\frac c {1+c} \le \frac {a+b} {1+c}\le \frac {a+b+2ab} {1+c}$  So the whole thing reduces to showing $c \ge a+b+ab$ ?  Can anyone help me with this, I'm rubbish at inequalities...
72180
This is a classic problem, but I haven't seen it on the site before.     Suppose $x$ and $y$ are real numbers that both lie in the interval   $(0,1)$. Prove  $$x^y+y^x&gt;1.$$   The following spoiler box contains a hint that indicates the solution I know. Does anyone know a more natural solution?       Use Bernoulli's inequality to show $$x^y\ge \frac{x}{x+y}.$$   Edit: I recently learned the source is the 1996 French math olympiad.
121303
Solving $\frac{1}{x} &lt; 4$ gives me $x &gt; \frac{1}{4}$. The book however states the answer is: $x &lt; 0$ or $x &gt; \frac{1}{4}$.  My questions are:  Why does this inequality has two answers (preferably the intuition behind it)?  When using Wolfram Alpha it gives me two answers, but when using $1 &lt; 4x$ it only gives me one answer. Aren't the two forms equivalent?
62490
Suppose $x,y,z$ are real positive numbers, prove that: $$\dfrac{x}{y} +\dfrac{y}{z} + \dfrac{z}{x} \ge 3$$ with equality when $x=y=z$.  Can someone help me find an easier solution? I started with assume only two are equal, without loss of generality, $x=y$. Then we get $$1 + \dfrac{x}{z} + \dfrac{z}{x}$$ $$1 + \dfrac{x^2 + z^2}{xz} = 1+ \dfrac{(x-z)^2 + 2xz}{xz} = 1 + 2 + \dfrac{(x-z)^2}{xz} \ge 3$$ To prove the general result I considered using a derivative with respect to $y$ to show that, starting with $x=y$ and increasing $y$ by a tiny amount, will show the equation's derivative as positive. I haven't gotten this to work yet. Any ideas?
11445
Prove that if $a&gt;0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $.   Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By factoring $(a^{10}-1)(a-1) \geq 0 $.  I am not sure how to proceed any further.