After what feels like an embarrassing hour of scribbling I can't seem to find a direct solution to the following problem  $Show \space that: a^2 + b^2 + c^2 \geq ab+bc+ca \space \space  \forall [a,b,c] \in Z^{+}_0 $  I've tried placing each integer in an arbitrary order like so:   $a\leq b \therefore ab \leq b^2$  $b \leq c \therefore bc \leq c^2 \implies$   $a \leq c \therefore$ $ac \leq c^2$  Naturally I tried to add up the inequalities but this clearly yielded no results, but by nature of the third inequality I run into issues; what have I missed?  EDIT  I've constructed newer perhaps more insightful inequalities from one of the Dr's answer below:  $a-b \leq ab \leq b^2$  $b-c \leq bc \leq c^2$  $a-c \leq ac \leq c^2$  EDIT 2  While I've seen this flagged as a possible duplicate, this question appears towards the beginning of an intro book to mathematical proofs without any prior knowledge given, I feel aso though this should be provable using pure inequalities from first principals
12190
How can the following inequation be proven?  $$a^2 + b^2 + c^2 \ge ab + bc + ca$$
8358
In the pdf which you can download  I found the following inequality which I can't solve it.     Exercise 2.1.11 Let $a,b,c  \gt 0$. Prove that    $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}.$$   Thanks :)
60825
Let $a$, $b$, $c$, $d$ and $e$ be positive numbers. Prove that:  $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+e}+\frac{e}{e+a}\geq\frac{a+b+c+d+e}{a+b+c+d+e-3\sqrt[5]{abcde}}$$  Easy to show that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ is true  and for even $n$ and positives $a_i$ the following inequality is true.  $$\frac{a_1}{a_1+a_2}+\frac{a_2}{a_2+a_3}+...+\frac{a_n}{a_n+a_1}\geq\frac{a_1+a_2+...+a_n}{a_1+a_2+...+a_n-(n-2)\sqrt[n]{a_1a_2...a_n}}$$
19100
Let $a,b,c,x,y,z&gt;0$ and $x+ y+ z= 1$. Show that $$ax+ by+ cz\geq  a^{x}.b^{y}.c^{z}$$ I have found the generalization of this inequality: For 2 sets of numbers $\left \{ a_{i} \right \}, \left \{ b_{i} \right \}&gt; 0$ and $\sum_{1}^{n}b_{i}= 1$, we have: $$\sum_{1}^{n}\left ( a_{i}b_{i} \right )\geq  \prod_{1}^{n}\left ( a_{i}^{b_{i}} \right )$$ First solution Simply compare the logs: as the natural logarithm is a concave function, one has $$\ln\bigl(a^x b^y c^z)=x\ln a+y\ln b+z\ln c\le \ln(ax+by+cz)$$ for all non-negative $x,y,z$ such that $x+y+z=1$. How about another solution? I really hope to see that. Thanks!
61013
How to prove for any complex numbers $a$, $b$, $c$, the inequality $$|a|+|b|+|c|+|a+b+c| \geq  |a+b|+|b+c|+|c+a|$$ is correct?
115559
I have been trying to show these two inequalities hold for all positive integers n, but I don't know how to proceed at all... I have tried playing around with them but I didn't find anything helpful. I have also tried induction but I couldn't make it work.
120544
Prove that if a,b.c are distinct positive integers that $$a^4+b^4+c^4&gt;abc(a+b+c)$$ My attempt:  I used the inequality A.M>G.M to get two inequalities First inequality $$\frac{a^4+b^4+c^4}{3} &gt; \sqrt[3]{a^4b^4c^4}$$  or $$\frac{a^4+b^4+c^4}{3} &gt; abc \sqrt[3]{abc}$$ or new -- first inequality $$\frac{a^4+b^4+c^4}{3abc} &gt; \sqrt[3]{abc}$$  second inequality: $$\frac{a+b+c}{3} &gt; \sqrt[3]{abc}$$ I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??
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.
113438
Let $a,b,c,d$ be positive real numbers. Prove that $$\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c} \geq \dfrac{2}{3}.$$   I was thinking of trying $a \geq b \geq c$ since the inequality is cyclic. I am not sure how to use this, though, or if this would help simplify the inequality. It also doesn't look like I can really use AM-GM or Cauchy-Schwarz.
38821
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.   My (strange) proof:  $$ \begin{align*} a^3+b^3+c^3 &amp;\geq a^2b+b^2c+c^2a\\ \sum\limits_{a,b,c} a^3 &amp;\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a^2 &amp;\geq \sum\limits_{a,b,c} ab\\ a^2+b^2+c^2 &amp;\geq ab+bc+ca\\ 2a^2+2b^2+2c^2-2ab-2bc-2ca &amp;\geq 0\\ \left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &amp;\geq 0 \end{align*} $$  Which is obviously true.    However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:  $$ \begin{align*} \sum\limits_{a,b,c} a^3 &amp;\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a &amp;\geq \sum\limits_{a,b,c} b\\ a+b+c &amp;\geq a+b+c \end{align*} $$  Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?  Edit: Please don't help me with the original inequality, I'll figure it out.
15491
Let $a,b,c$ are positive number such that $abc=1$. Prove that:   $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\;\geq\; a+b+c$$   This problem froms my Math teacher. I have attempted to let $$(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$$. The inequality is equivalent to: $\frac{\sqrt{x^2-xy+y^2}}{y}+\frac{\sqrt{y^2-yz+z^2}}{z}+\frac{\sqrt{x^2-xz+z^2}}{x}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$  Then, I tried to use AM-GM but I stucked on it.
103235
$$(a_1 + a_2 + · · · + a_n)( 1/a_1 +1/a_2+ · · · +1/a_n) ≥ n^2$$  My TA mentioned he solved this using Cauchy Schwartz inequality, but I don't see how. All I've managed to do is make a summation of $1$, $n$ times, which ends up being $n ≥ n^2$.
14394
Prove that  $(x_1+\dots+x_n)^2 \leq n(x_1^2 + \dots+x_n^2)$  for all positive integers n and all real numbers $x_1,....,x_n$  I am attempting a proof by induction but wasn't sure if i need the Cauchy-Schwarz Inequality or perhaps another way other than induction to prove this.  Proof   $n=1$ true  assume true for $n=k$  Now for $n= k+1$  $(x_1 + \dots +x_k + x_{k+1})^2 \leq \dots$
74202
I have been working on the following problem:     For a given $c\in \Bbb Z^+$, find $a,b \in \Bbb Z^+: c=a^2-b^2$.   I have already figured out and proved a number of things:   A way to directly determine if a particular instance is solvable:  $\exists a,b \in \Bbb Z^+: c=a^2-b^2 \iff c \not\equiv 2 \mod 4$ An upper bound on the values for a (and b): $b \lt a\le\lfloor\frac{c-1}{2}\rfloor$ And a simple algorithm to find a solution:     $$ \begin{align*}  (a_0, b_0) &amp;= (0, 0) \\  (a_{n+1}, b_{n+1}) &amp;= \begin{cases}   (a_n+1, &amp;b_n&amp;)&amp;\text{if } a_n^2-b_n^2 &lt; c\\   (a_n, &amp;b_n+1&amp;)&amp;\text{if } a_n^2-b_n^2 &gt; c\\  \end{cases} \end{align*} $$   This can be iterated until a solution is found or the upper bound is reached.   I am currently trying to work out a more direct way to find an answer, but I have been unable to figure out any other ways to approach the problem.  How can I more directly find values for $a$ and $b$ for a given value of $c$?
81791
Determine the set of points $z$ that satisfy the condition $|2z|&gt;|1+z^2|$  I tried to redo this problem and got to this point  $|2z|&gt;|1+z^2|$ $\Rightarrow$ $2|z|&gt;1+|z^2|$ $\Rightarrow$ $2|z|&gt;1+z\overline z$  Let $z=x+iy$ then   $$2\sqrt{x^2+y^2}&gt;1+(x+iy)(x-iy)$$  $$2\sqrt{x^2+y^2}&gt;1+(x^2+y^2)$$  $$0&gt;1-2\sqrt{x^2+y^2}+(x^2+y^2)$$  $$0&gt;(1-\sqrt{x^2+y^2})^2$$  since $x,y$ are real number , so there is no solution for this inequality?
36372
I need to use $|a+b| \leq |a|+|b|$ to show that $||a|-|b|| \leq |a-b|$ .   I have tried to represent  $||a|-|b||$ as  $||a|+(-|b|)|$ , and then get $||a|+(-|b|)| \leq |a|+|-|b||$ , but that isn't leading me anywhere given $|a-b| \leq |a|+|b|$.
43018
Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.  Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$    Here's my idea:  $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  And I'm stuck here.  I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$  In the first case using Cauchy-Schwarz Inequality I prove that:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $3(a^2 + b^2 + c^2) \ge (a+b+c)^2$  Now I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$   $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$   I need I don't know how to continue.  In the second case I tried proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and  $a^2 + b^2 + c^2 \ge a+b+c$  Using Cauchy-Schwarz Inequality I proved:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$  $a^2 + b^2 + c^2 \ge a+b+c$  But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$  So please help me with this problem.  P.S   My initial idea, which is proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  maybe isn't the right way to prove this inequality.