Questions tagged [buffalo-way]

Inequalities that can be proved by BW (Buffalo Way).

  • For example:

Prove that $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0$$ for non-negatives $x$, $y$ and $z$.

A proof by BW:

Let $x$ be a minimal number between numbers $x$, $y$ and $z$, $y=x+u$ and $z=x+v$.

Hence, $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)=(u^2-uv+v^2)x+(u+v)(u-v)^2\geq0.$$

Sometimes it is better to use $y=x+u$, $z=x+u+v$ or $y=x+u+v$, $z=x+u$,

where $u$ and $v$ are non-negatives.

  • If $x$, $y$ and $z$ are sides-lengths of triangle and $z=\max\{x,y,z\}$, we can use the following substitution:

$x=a+u$, $y=a+v$ and $z=a+u+v$, where $a>0$, $u\geq0$ and $v\geq0$.

For example:

For all triangle prove that: $$a^3b^2+b^3c^2+c^3a^2\geq(a^2+b^2+c^2)abc$$

Proof:

Let $c$ be a maximal number between numbers $a$, $b$ and $c$, $a=x+u$, $b=x+v$ and $c=x+u+v$, where $x>0$, $u$ and $v$ are non-negatives.

Hence, $$a^3b^2+b^3c^2+c^3a^2-(a^2+b^2+c^2)abc=(u^2-uv+v^2)x^3+$$ $$+3(u^3-uv^2+v^3)x^2+(3u^4+2u^3v-3u^2v^2-2uv^3+3v^4)x+u^5+u^4v-2u^2v^3+v^5\geq0.$$

  • Another example.

For non-negatives $a$, $b$, $c$ and $d$ prove that: $$a^4+b^4+c^4+d^4+4abcd\geq2(a^2bc+b^2cd+c^2da+d^2ab).$$

A proof:

Let $a=\min\{a,b,c,d\}$, $b=a+u$, $c=a+v$ and $d=a+w$.

Hence, $$a^4+b^4+c^4+d^4+4abcd-2(a^2bc+b^2cd+c^2da+d^2ab)=$$ $$=2(2u^2+2v^2+2w^2-uv-uw-vw)a^2+$$ $$+2(2u^3+2v^3+2w^3-u^2v-u^2w-v^2w-w^2u)a+u^4+v^4+w^4-2u^2vw\geq0$$

150 questions
48
votes
5 answers

For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)

Let $a$ , $b$ , $c$ , and $d$ be the lengths of the sides of a quadrilateral. Show that $$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$ Background: The well known 1983 IMO Problem 6 is the following: IMO 1983 #6. Let $a$ , $b$ and $c$…
31
votes
6 answers

Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$

Suppose that $a,b,c,x,y,z$ are all positive real numbers. Show that $$\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$$ Below are what I've done, which may be misleading. I've tried to analyze when the equality…
Mathis
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15
votes
2 answers

If $a+b+c+d=1$ so $\sum\limits_{cyc}\sqrt{a+b+c^2}\geq3$

Let $a$ , $b$ , $c$ and $d$ be non-negative numbers such that $a+b+c+d=1$ . Prove that: $$\sqrt{a+b+c^2}+\sqrt{b+c+d^2}+\sqrt{c+d+a^2}+\sqrt{d+a+b^2}\geq3.$$ I tried C-S, Holder and more, but without success. For example, we need to prove that:…
13
votes
4 answers

Given triangle side lengths $a, b, c$, show that $3\left(a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\right)\geq b\left(a+b-c\right)\left(a-c\right)\left(c-b\right)$

If you're interested in IMO 1983-style inequalities, please consider the following problem : Given three positive real numbers $a, b, c$ that form the side lengths of a triangle, prove inequality: $$3\left ( a^{2}b\left ( a- b \right )+ b^{2}c\left…
13
votes
3 answers

Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

The inequality: $$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$ Conditions: $a,b,c,d \in \mathbb{R^+}$ I tried using the normal Cauchy-Scharwz, AM-RMS, and…
12
votes
2 answers

A cyclic inequality $\sum\limits_{cyc}{\sqrt{3x+\frac{1}{y}}}\geqslant 6$

Given positive real numbers $x,y,z$ satisfying $x+y+z=3$ , prove that $$ \sqrt{3x+\frac{1}{y}}+\sqrt{3y+\frac{1}{z}}+\sqrt{3z+\frac{1}{x}}\geqslant 6. $$
Steven Sun
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11
votes
2 answers

cyclic three variable inequality

Let $a,b,c$ be nonnegative real numbers and $a+b+c=3$. Prove the inequality $$ \sqrt{24a^2b+25}+\sqrt{24b^2c+25}+\sqrt{24c^2a+25}\le 21 $$ I have tried to find the solution using classical inequalities, but failed. Any idea?
10
votes
5 answers

Prove $\frac{a}{a+bc}+\frac{b}{b+cd}+\frac{c}{c+da}+\frac{d}{d+ab}\ge 2$ for positives $\sum a = 4$

Question: Let $$a,b,c,d>0,a+b+c+d=4$$ show that $$\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$ when I solved this problem, I have see following three variables inequality: Assumming that $a,b,c>0,a+b+c=3$ , show that :…
math110
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9
votes
3 answers

Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+5b^2}\geq\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}$ for reals

Problem . Let $a, b, c \in \mathbb{R}$ . Prove that $$ \sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geqslant\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}. $$ I think that this inequality is strong, though I do not have knowledge of many techniques. There…
9
votes
2 answers

Proving $\sum\limits_{\text{cyc}} \frac{a}{b^2+c^2+d^2} \geq \frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}$ for $a, b, c, d >0$

Let $a, b, c, d > 0$ . Prove that $$\frac{a}{b^2+c^2+d^2}+\frac{b}{a^2+c^2+d^2}+\frac{c}{a^2+b^2+d^2}+\frac{d}{a^2+b^2+c^2}$$ $$\geq\frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}.$$ What I tried, was to say that $a^2+b^2+c^2+d^2=1$ and so…
9
votes
6 answers

Prove that $\sum\limits_{cyc}\frac{a}{a^2+ab+b^2+3}\leq\frac{1}{2}$

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$\frac{a}{a^2+ab+b^2+3}+\frac{b}{b^2+bc+c^2+3}+\frac{c}{c^2+ca+a^2+3}\leq\frac{1}{2}$$ I think this inequality is very interesting because most of the contest's inequalities are homogeneous,…
9
votes
6 answers

For any triangle with sides $a$, $b$, $c$, show that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$

For any triangle with sides $a$ , $b$ , $c$ , prove the inequality $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0 .$$ This is IMO 1983 problem 6. I tried substituting $a=x+y$ , $b=y+z$ , $c=z+x$ but well it doesn't help in any sense except wasting 3 pages…
8
votes
1 answer

Why is the "Buffalo Way" considered inelegant?

I was going through an "article" on the "Buffalo Way" , where the author said that one should NEVER use the Buffalo Way for proving inequalities in actual real-time contests as it is "highly inelengant". What is the reason behind this notion ? In…
user399078
7
votes
0 answers

Prove $abc (a^{2} + bc + ca)( b^{2} + ca + ab)(c^{2} + ab + bc ) \le 27$ subject to $(4a + 5b)(4b + 5c)(4c + 5a) = 729$

Let $a, b, c > 0$ such that $\left( 4a + 5b \right) \left( 4b + 5c \right) \left( 4c + 5a \right) = 729$ . Prove that $$abc \left( a^{2} + bc + ca \right) \left( b^{2} + ca + ab \right) \left( c^{2} + ab + bc \right) \le 27.$$ This question was…
7
votes
2 answers

Find maximum $k \in \mathbb{R}^{+}$ such that $ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $

Find maximum $k \in \mathbb{R}^{+}$ such that $$ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $$ for all $a, b, c$ that are distinct positive real numbers ( $a \neq b$ , $b \neq c$ , $a \neq c$ ) Usually when I see…
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