Linked Questions

3
votes
6answers
259 views

Prove that $xy+yz+zx \leq x^2+y^2+z^2$ [duplicate]

Prove that $xy+yz+zx \leq x^2+y^2+z^2$ . Hint: Use $\frac{a+b}{2}\geq\sqrt{ab}$ First I tried using the hint by setting $a=x$ and $b=y+z$, however this results in the inequality: $$x^2+y^2+z^2 \geq ...
6
votes
5answers
168 views

Finding the minimum value of $a^2+b^2+c^2$ [duplicate]

Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$. I've been trying to solve this, but I don't really know how to approach this. I ...
5
votes
4answers
473 views

Proving the following quadratic inequality? [duplicate]

Apologies if this has been asked before - I could not find a question with this exact inequality. Basically the inequality is $$(a+b+c)^2 \leq 3 a^2 + 3 b^2 + 3 c^2$$ Expanding it out we see that $...
2
votes
2answers
1k views

Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$ [duplicate]

Show that $a^2 + b^2 + c^2 \geq ab + bc + ca$ for all positive integers $a$, $b$, and $c$. I am not sure how to approach this problem. Should I divide this problem into multiple cases based on ...
2
votes
3answers
112 views

Simple two variable am-gm inequality [duplicate]

Given $x,y \in \Bbb{R}$, show that:$$x^2+y^2+1\ge xy+y+x $$ I tried using the fact that $x^2+y^2 \ge 2xy$ But then I'm not sure how to go on, Also tried factoring but didn't help much, also tried ...
0
votes
3answers
116 views

How do I prove this inequality: $a^2+b^2+c^2 \geq ab+bc+ac$? [duplicate]

How do I prove that for any $a, b, c \in \mathbb{R}$ the inequality $a^2+b^2+c^2 \geq ab+bc+ac$ is true?
2
votes
3answers
155 views

Proof help: Prove that $x^2+y^2+z^2 \geq xy+xz+yz$ [duplicate]

$x^2+y^2+z^2 \geq xy+xz+yz $ for all real numbers, x, y, and z. I'm not very good with working inequality proofs. Can someone help me prove this? The technique doesn't really matter.
1
vote
2answers
64 views

How to explain this inequality in a direct way? [duplicate]

$a^2 +b^2 +c^2 \ge a(b+c)+bc$, this equation is supposed to be correct, although I factorized and tried to solve prove in many ways, I cant explain well enough!
0
votes
2answers
64 views

Proving a seemingly simple inequality is proving difficult [duplicate]

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]...
1
vote
4answers
19k views

Is this correct method to prove that $a^2 + b^2 + c^2 ≥ ab + bc + ac$, when $a,b,c \geq 0$?

Can I prove it like this: Let's say that $a=b=c$ so we get "If $a \geq 0$ then $3a^2 ≥ 3a^2$" Now I take the negation of that statement and get "If $a \geq 0$ then $3a^2 < 3a^2$" The anti-thesis is ...
7
votes
3answers
470 views

Geometric inequality: $2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$

Suppose $a$, $b$, and $c$ are the lengths of the sides of a triangle, and $R$ and $r$ are its circumradius and inradius respectively. How can one prove the following inequality? $$2r^2+8Rr \leq \frac{...
-3
votes
4answers
172 views

Prove the inequality $a^2 + b^2 +c^2 \ge ab +bc +ac$ [closed]

How do I prove the inequality \begin{equation*} a^2 + b^2 +c^2 \ge ab +bc +ac \end{equation*} where $ a,b,c\in\mathbb{R} $ and $a,b,c>0$? I obtained only $(a+b+c)^2\ge 3(ab+bc+ac)$ and some ...
1
vote
4answers
98 views

Prove that: $\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$

Prove that for nonegative $x,y,z$ we have: $$\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$$ I prove that using the tangent line method. We may assume that $x+y+z=1$, so you we have to ...
1
vote
0answers
165 views

Prove that $\sum\limits_{cyc}\sqrt[4]{a^2+bc+ca}\geq3\sqrt[4]{ab+ac+bc}$

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$\sqrt[4]{a^2+bc+ca}+\sqrt[4]{b^2+ca+ab}+\sqrt[4]{c^2+ab+bc}\geq3\sqrt[4]{ab+ac+bc}$$ I tried Holder and more, but without any success. I ...