The lengths $a,b$ and $c$ are the side lengths of a triangle that satisfy $ab+bc+ca=3$. Prove that $3≤a+b+c≤2√3$  I was able to find the minimum bound through some simple algebra: $a+b+c=\sqrt{a^2+b^2+c^2+6}$ But, $a^2+b^2+c^2≥ab+bc+ca=3$ $\implies a+b+c≥\sqrt{3+6}$ $$3≤a+b+c$$ Alternatively we could use AM GM on $ab+bc+ca$ and then on $a^2+b^2+c^2$ After this I couldn't think of how to use the triangle inequality in conjunction with the given condition $ab+bc+ca=3$ to get an upper bound on the sum, any help would be appreciated, thanks
122765
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.    The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequality. How do I show the other inequality?
46034
Given $a,b,c \in \mathbb{R^+}$ such that $a+b+c=12$   Find Minimum value of $$S=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2$$  My Try: By Cauchy Schwarz Inequality we have  $$\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{c}\right)+\left(c+\frac{1}{a}\right)\le \sqrt{3}\sqrt{S}$$  $\implies$  $$\sqrt{3S} \ge 12+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$  Now by $AM \ge HM $ inequality we have  $$\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ $\implies$  $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{3}{4}$$  hence  $$\sqrt{3S} \ge 12+\frac{3}{4}=\frac{51}{4}$$  hence  $$3S \ge \frac{2601}{16}$$  so $$S \ge \frac{867}{16}$$  is this approach correct and any better approach please share.
81975
Let $a,b \in [0,1], a+b=1$ and let $\{u,v\}\subset\mathbb{R}$.   How can we prove that $$(au+bv)^2\leq au^2+bv^2?$$   Opening the brackets give me in the left side $(au+v-av)^2$, but I'm not think that's the right way to do that.  I don't find the triangle inequality very helpfull, too.  Thanks.
22888
Prove that in any triangle $ABC$ the following inequality holds $$\cos{A}+\cos{B}+\cos{C}\ge\dfrac{1}{4}(3+\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$  And I have gotten $$8(\cos{A}+\cos{B}+\cos{C})\ge 6+2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$  $$2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})+3=(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$ $$\Longleftrightarrow 8\sum_{cyc}\cos{A}\ge 3+(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$ then Any hints, ideas? Thanks in advance.
27964
Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality :   $$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$  Where :   $s : \operatorname{Semiperimeter}$  $R : \operatorname{circumradius}$   I know the relation :   $$m_{a}^{2}=\frac{2(b^{2}+c^{2})-a^{2}}{4}$$  But when I multiple together I dont get simple formulas!  So, I need help finding a solution. Thanks!
74826
Show that $\max{\{|a|+|b|,|c|+|d|\}} \leq \max{\{|a|,|c|\}}+\max{\{|b|,|d|\}}.$  I wanted to show that $d(p,q)=\max{\{|x_1-x_2|,|y_1-y_2|\}}$ where $p=(x_1,y_1),q=(x_2,y_2)$ is a metric on $\mathbb{R^2}.$ In proving the triangle inequality let $p=(x_1,y_1),q=(x_2,y_2),r=(x_3,y_3).$ Then  \begin{align*} d(p,r)&amp;=\max{\{|x_1-x_3|,|y_1-y_3|\}}\\ &amp;\le \max{\{|x_1-x_2|+|x_2-x_3|,|y_1-y_2|+|y_2-y_3|\}}\\ &amp;\le \max{\{|x_1-x_2|,|y_1-y_2|\}}+\max{\{|x_2-x_3|,|y_2-y_3|\}}=d(p,q)+d(q,r) \end{align*}  I don't understand how this inequality is true.
118215
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 &lt; 2(xy + yz + xz)$        My solution: Consider  $$x^2 + y^2 + z^2 &lt; 2(xy + yz + xz)$$  Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$  Hence  $$(x+y+z)^2-2(xy+yz+xz) &lt; 2(xy + yz + xz)$$  $$ (x+y+z)^2 - 4(xy+yz+xz) &lt; 0 $$  As $x,y,z$ are all greater than zero as a side length of a triangle can not be negative. And because $(x+y+z)^2 &gt;0$ for all real $x,y,z$ therefore the whole expression is less than zero. $Q.E.D$  Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :)
70242
I've tried solving $\sqrt[3]{3} =a + b* \sqrt[3]{2}+c* \sqrt[3]{4}$, but there is no obvious contradiction, even when taking the norms/traces of both sides. I can't think of another approach.  This is part of a proof, it seems obvious, but it's really bothersome I can't prove it.
48670
$a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$    My work: I tried using Cauchy-Schwarz inequality to find that, $(a^2+b^2+c^2)(1^2+1^2+1^2)\ge (a+b+c)^2$ $(a^2+b^2+c^2)\ge \dfrac13(a+b+c)^2$ $\sqrt{(a^2+b^2+c^2)}\ge \dfrac{1}{\sqrt{3}}(a+b+c)\ge \dfrac{1}{\sqrt{3}}abc$ which is not what I need and neither I can use it to prove the required inequality. Please help.
13874
If $\triangle$ be the area of $\triangle ABC$ with side lengths $a,b,c$. Then show that $\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}$   and also show that equality hold if $a=b=c$.  $\bf{My\; Try}::$ Here we have to prove $4\triangle\leq \sqrt{(a+b+c)\cdot abc}$  Using the formula $$\triangle = \sqrt{s(s-a)(s-b)(s-c)},$$ where $$2s=(a+b+c)$$  So $$4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$  Now using $\bf{A.M\geq G.M}$ for $(b+c-a)\;,(c+a-b)\;,(a+b-c)&gt;0$  $$\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$$  So we get $\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}$  But I did not understand how can I prove above inequality  help Required  Thanks
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 :)
116465
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$ then proving $\triangle$ is equilateral.  $\bf{My\; Try::}$ We can write it as $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1$$ So $$\left(\frac{a+b-c}{a}\right)^{a}\cdot \left(\frac{b+c-a}{b}\right)^{b}\cdot \left(\frac{c+a-b}{c}\right)^{c}\geq 1$$ So $$(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$$ Now how can i solve after that, Help required, Thanks
119929
Let $a,b,c\in\Bbb R^+$. For any given $n\in\Bbb N$ there exists a triangle with sides $a^n,b^n,c^n$. Show that at least two of $a,b,c$ are equal.   I know that you can create inequalities for $a^n,b^n,c^n$, but what's next?
121932
We have an acute triangle with angles $\alpha, \beta, \gamma$ and we need to tell whether the following inequality is true for all such triangles:  $$(\tan(\frac{\alpha}{2}))^2+(\tan(\frac{\beta}{2}))^2+(\tan(\frac{\gamma}{2}))^2\geq1$$  We are also asked if $1%$ is the greatest number, for which it is true.  I'm having huge problems when trying to solve such optimization calculus problems. I think we must somehow find triangle for which we can prove, that the upper sum is minimum, than evaluate it and prove that it is equal to $1$. As with most optimization problems, it turns out that if $\alpha=\beta=\gamma=\frac{\pi}{3}$, then the sum evaluates to $1$. But how to prove that this is infimum of this sum?
32334
This is supposed to be an application of AM-GM inequality.     if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$   First of all,  $a^2+b^2+c^2\ge 3$  by a direct application of AM-GM.Also,we have  $a^2+b^2+c^2\ge ab+bc+ca$  Next,we consider the expression  $(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$  but that hardly helps.I know that  $3(a^2+b^2+c^2)\ge (a+b+c)^2$  From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,  $a^3+b^3+c^3\ge a^2b+b^2c+c^2a$  $(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$ A hint will be appreciated at this point.
17339
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.   I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing  helps.  On this topic we learned also about Cauchy inequality, but I have no experience with it.  The answer according to Mathematica is when $A=B=C=60$.  Any ideas?
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