WLOG $a=\min(a,b,c)$ and set $b=a+x, c=a+y$. We then clearly have \[\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}=\frac{a^2}{(x-y)^2}+\frac{(a+x)^2}{y^2}+\frac{(a+y)^2}{x^2}\ge\frac{x^2}{y^2}+\frac{y^2}{x^2}>2\]

InCtrl wrote: For pairwise distinct nonnegative reals $a,b,c$, prove that $$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}>2$$. It's so easy! Let$a>b>c$ and $f(a,b,c)=\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}-2$ It's very easy to see that $$f(a,b,c)\ge f(a-t,b-t,c-t)\ \ \ \text{where}\ \ \ c\ge t\ge0$$so set $t=c$ then we have $$f(a,b,c)\ge f(a,b,0)=\frac{a^2}{b^2}+\frac{b^2}{a^2}-2>0\ \ \ \text{So easy by AM-GM}$$GOOD LUCK

InCtrl wrote: For pairwise distinct nonnegative reals $a,b,c$, prove that $$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}>2$$. The following inequality is also true The real numbers $a,b,c $ are distinct. Prove that\[\left |\frac{a}{b-c}\right |+ \left |\frac{b}{c-a}\right |+ \left |\frac{c}{a-b}\right |\ge2 .\]

InCtrl wrote: For pairwise distinct nonnegative reals $a,b,c$, prove that $$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}>2$$. Proof: WLOG $a>b>c\geq 0.$ We have$$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2}\geq \frac{a^2}{(b-c)^2}+\frac{b^2}{(a-c)^2}\geq 2\cdot \frac{a}{a-c}\cdot \frac{b}{b-c}>2\cdot 1\cdot 1=2$$https://artofproblemsolving.com/community/c6h37317p234720 Middle school mathematics(China Wuhan)No.9(2008)

Does there exist a solution with a) Cauchy or b) Muirhead?

Probably no solution with Cauchy given that the equality cases are different. $a=0,b=1,c=1$ is an equality case for this if we ignore pairwise distinct, which is never a cauchy equality case. I may be wrong, though, so feel free to correct me.

The equation is symmetric, so let $a>b>c$. Let $f(c)=\frac{a^2}{(b-c)^2}+\frac{b^2}{(a-c)^2}$. When $f(0)$, one obtains $\frac{a^2}{b^2}+\frac{b^2}{a^2}$. Since $a\neq b$, $f(c)$ is greater than $2$, by AM-GM. Notice that as $c$ approaches from $0$ to $b$, the denominators decrease, meaning that $f(c)$ is a strictly increasing function.

Coincidentally

My solution: Note that if we put $x=\frac{a}{b-c} , y=\frac{b}{c-a} , z=\frac{c}{a-b}$ Then we have $\sum xy=\frac{a^2b-ab^2+b^2c-bc^2+c^2a-a^2c}{(a-b)(b-c)(c-a)}=-1$ And we know that if $\sum xy<0$ then $\sum x^2\geq -2\sum xy$ So done!

L.H.S.=$\sum\frac{a^4}{(ab-ac)^2} \ge \frac{(a^2 +b^2 + c^2)^2}{2(a^2b^2 + a^2c^2 + b^2c^2 - abc(a+b+c))} \ge 2 $ if and only if $ a^4 + b^4 +c^4 +4abc(a+b+c) \ge 2(a^2b^2 + a^2c^2 + b^2c^2)$. But $a^4 +b^4 +c^4 + 3abc(a+b+c) \ge \sum ab(a^2+b^2) \ge 2\sum a^2b^2$ (Schur and AM GM) and since $abc(a+b+c) > 0$, we are done.

Cool problem! Suppose $\exists \, x,y,z \in \{a,b,c\} $ such that $x+y \leq z$. Then exactly two of $ \left \{ \frac{a^2}{(b-c)^2}, \frac{b^2}{(c-a)^2}, \frac{c^2}{(b-a)^2} \right \}$ are $\geq 1$ (equality can't hold in both cases) while the third expression is nonnegative. The inequality thus follows. Otherwise, we have $x+y > z$, which implies that $a,b,c$ are sides of a triangle. Then use Cauchy to get $$LHS \geq \frac{(a+b+c)^2}{2a^2+2b^2+2c^2-2ab-2bc-2ca} > 2$$where the RHS of the last inequality follows from $$2ab+2bc+2ca > a^2 + b^2 + c^2$$which holds true for $a,b,c$ sides of a triangle (you may use Ravi substitution).

InCtrl wrote: For pairwise distinct nonnegative reals $a,b,c$, prove that $$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}>2$$. I have to say,it'an old inequality by Vasile Cirtoaje: $\frac{{{a^2}}}{{{{(b - c)}^2}}} + \frac{{{b^2}}}{{{{(c - a)}^2}}} + \frac{{{c^2}}}{{{{(b - a)}^2}}}$ $ = {\left( {\frac{a}{{b - c}} + \frac{b}{{c - a}} + \frac{c}{{b - a}}} \right)^2} - 2\left( {\frac{{ab}}{{\left( {b - c} \right)\left( {c - a} \right)}} + \frac{{bc}}{{\left( {c - a} \right)\left( {a - b} \right)}} + \frac{{ca}}{{\left( {a - b} \right)\left( {b - c} \right)}}} \right)$ $ = {\left( {\frac{a}{{b - c}} + \frac{b}{{c - a}} + \frac{c}{{b - a}}} \right)^2} + 2$

Hermitianism wrote: InCtrl wrote: For pairwise distinct nonnegative reals $a,b,c$, prove that $$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}>2$$. I have to say,it'an old inequality by Vasile Cirtoaje: $\frac{{{a^2}}}{{{{(b - c)}^2}}} + \frac{{{b^2}}}{{{{(c - a)}^2}}} + \frac{{{c^2}}}{{{{(b - a)}^2}}}$ $ = {\left( {\frac{a}{{b - c}} + \frac{b}{{c - a}} + \frac{c}{{b - a}}} \right)^2} - 2\left( {\frac{{ab}}{{\left( {b - c} \right)\left( {c - a} \right)}} + \frac{{bc}}{{\left( {c - a} \right)\left( {a - b} \right)}} + \frac{{ca}}{{\left( {a - b} \right)\left( {b - c} \right)}}} \right)$ $ = {\left( {\frac{a}{{b - c}} + \frac{b}{{c - a}} + \frac{c}{{b - a}}} \right)^2} + 2$ I agree i am pretty sure that i saw this inequaity in Vasile book Algebraic Inequalities chapter 1 page 5 problem 6

Titu lemma will kill this as #10

Suppose that $a>b>c$ and let $b=c+y$, $a=b+x=c+x+y$. Then we have \begin{align*} LHS =& \frac{(c+x+y)^2}{y^2} +\frac{(c+y)^2}{(x+y)^2} +\frac{c^2}{x^2}\\ \geqslant & \frac{(c+x+y)^2}{y^2} +\frac{(c+y)^2}{(x+y)^2} \\ \geqslant & \frac{(x+y)^2}{y^2} +\frac{y^2}{(x+y)^2} >2. \end{align*}

Why is this so simple?

Is there a similar ineqality with four variables a,b,c,d, non négative?

I get that the expression is greater than 3. Using AM-GM I get that the expression is minimised when $\frac{a}{b-c} =\frac{b}{c-a}=\frac{c}{b-a}=1$ The rest follows

Mathcat1234 wrote: I get that the expression is greater than 3. Using AM-GM I get that the expression is minimised when $\frac{a}{b-c} =\frac{b}{c-a}=\frac{c}{b-a}=1$ The rest follows You are wrong

Really cool problem! Would like to see more inequalities using ideas like this. WLOG $a>b>c$. Consider the function $f(c) = \frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}$, where $a$ and $b$ are fixed. Note that as $c$ increases, $f(c)$ increases, so the minimum of this function is at $c=0$. But then $f(0) = \frac{a^2}{b^2} + \frac{b^2}{a^2} = \frac{a^4+b^4}{a^2b^2} > 2$, since $a>b$.

Dang how's smoothing. Let $P(a, b, c)$ denote the 3-variable (assertion?). WLOG suppose $a > b > c$. Notice that $P(a, b, c) \geq P(a- c, b-c, 0)$ hence shifting all variables down by $c$ decreases the value. We may assume $c = 0$. AM-GM finishes. However, equality can never occur, so the sign is strict. Remark: Can't believe I wasted my time typing this up.

sqing wrote: InCtrl wrote: For pairwise distinct nonnegative reals $a,b,c$, prove that $$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}>2$$. Let $a,b,c$ be distinct real numbers. Then $$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(b-a)^2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}.$$(Vasile Cirtoaje) If $ a,b,c$ are nonnegative real numbers,then $ (a) \ \ \ \ \left(\frac a{b - c}\right)^2 + \left(\frac b{c - a}\right)^2 + \left(\frac c{a - b}\right)^2\ge \frac {a^2 + b^2 + c^2}{ab + bc + ca}$; $ (b) \ \ \ \ \left(\frac a{b - c}\right)^2 + \left(\frac b{c - a}\right)^2 + \left(\frac c{a - b}\right)^2\ge \frac {(a + b + c)^2}{2(ab + bc + ca)}$. Moreover, $ (c) \ \ \ \ \left(\frac a{b - c}\right)^2 + \left(\frac b{c - a}\right)^2 + \left(\frac c{a - b}\right)^2+2\ge \left(\frac {a^2 + b^2 + c^2}{ab + bc + ca}\right)^2$.

WLOG let $a < b < c$ and let $b=a+x$ and $c=a+y$. Then substituting we get \[ \frac{a^2}{(x-y)^2}+\frac{(a+x)^2}{y^2}+\frac{(a+y)^2}{x^2} > 2. \]The LHS is minimized when $a=0$ so we are left with \[ \frac{x^2}{y^2}+\frac{y^2}{x^2}>2 \]By AM-GM we get that \[ \frac{x^2}{y^2}+\frac{y^2}{x^2} \ge 2 \]Notice that the equality case is when $x=y$ but since $b \neq c$ this does not hold. Thus we are done.

Use BW with $c=\min\{a,b,c\}$ and $b=c+x$, $a=c+y$. Since $x\ne y$: $$\sum_{\text{cyc}}\frac{a^2}{(b-c)^2}=\frac{(c+y)^2}{x^2}+\frac{(c+x)^2}{y^2}+\frac{c^2}{(x-y)^2}\ge\frac{(c+y)^2}{x^2}+\frac{(c+x)^2}{y^2}\ge\frac{x^2}{y^2}+\frac{y^2}{x^2}>2.$$

Titu's lemma helps WLOG, we assume that $a\geq b\geq c$, rewrite the equation to $\frac{a^2}{(b-c)^2}+\frac{b^2}{(a-c)^2}+\frac{c^2}{(a-b)^2}\geq \frac{(a+b+c)^2}{(a-b)^2+(a-c)^2+(b-c)^2}$ Now we have to prove that $\frac{(a+b+c)^2}{(a-b)^2+(a-c)^2+(b-c)^2}\geq 2$, after expanding, we have to prove that $a^2+b^2+c^2\leq 2ab+2bc+2ac$, which is easy with AM-GM. However, the equal sign can happen and can only happen when $a=b=c$, as the problem said, they are not equal to each other, so the equal sign cannot hold so $\frac{a^2}{(b-c)^2}+\frac{b^2}{(a-c)^2}+\frac{c^2}{(a-b)^2}>2$ as desired

A correct math problem and nice solutions

Inequality is homogenous, so assume $a^2+b^2+c^2=1.$ Note that \begin{align*} \frac{a^2}{(b-c)^2}+\frac{b^2}{(a-c)^2}+\frac{c^2}{(a-b)^2} &\geq \frac{(a+b+c)^2}{(a-b)^2+(a-c)^2+(b-c)^2}\\ &=\frac{1}{2} \cdot \frac{a^2+b^2+c^2+2ab+2bc+2ca}{a^2+b^2+c^2-ab-bc-ca} \\ &= \frac{1}{2} \cdot \frac{1+2(ab+bc+ca)}{1-(ab+bc+ca)} \\ &= \frac{1}{2} \cdot \frac{1+2x}{1-x} \\ \end{align*}where we let $x=ab+bc+ca.$ Thus, it suffices to show \begin{align*} \frac{1}{2} \cdot \frac{1+2x}{1-x} &\geq 2\\ \frac{1+2x}{1-x} &\geq 4\\ 1+2x &\geq 4-4x \\ x &\geq \frac{1}{2} \\ 2(ab+bc+ca) \geq a^2+b^2+c^2.\\ \end{align*}However, we have that by AM-GM, $$\frac{a^2+b^2}{2} \geq ab$$$$\frac{b^2+c^2}{2} \geq bc$$$$\frac{a^2+c^2}{2} \geq ac,$$and summing proves the desired result.

samrocksnature wrote: However, we have that by AM-GM, $$\frac{a^2+b^2}{2} \geq ab$$$$\frac{b^2+c^2}{2} \geq bc$$$$\frac{a^2+c^2}{2} \geq ac,$$and summing proves the desired result. Add them can only get $a^2+b^2+c^2\geq ab+bc+ca$(did I miss something?)

whoops

You might not believe it, but this is simply a perfect square of a fraction! \[\text{left}-\text{right}=\left(\frac{\sum a(a-b)(a-c)}{\prod(a-b)}\right)^2>0.\]

Let $a,b,c\ge 0$ and $(a-2b)(b-2c)(c-2a)\neq 0.$ Prove that $$\frac{a^2}{(b-2c)^2}+\frac{b^2}{(c-2a)^2}+\frac{c^2}{(a-2b)^2}\ge 1$$

Here is an incredibly elegant proof: Define $$x = \frac{a}{b-c}, y = \frac{b}{c-a}, z = \frac{c}{a-b}$$Consider the following matrix \[ M= \left[ {\begin{array}{ccc} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \\ \end{array} } \right] \] This is introduced so as to satisfy the following condition: \[ M\left[ {\begin{array}{c} a \\ b \\ c \\ \end{array} } \right] = \left[ {\begin{array}{ccc} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \\ \end{array} } \right] \left[ {\begin{array}{c} a \\ b \\ c \\ \end{array} } \right] = \left[ {\begin{array}{c} a-bx+cx \\ ay+b-cy \\ -az+bz+c \\ \end{array} } \right] = \left[ {\begin{array}{c} 0 \\ 0 \\ 0 \\ \end{array} } \right] \] Now, if $\det(M) \neq 0$, then $M^{-1}$ such that $M^{-1}M = I$ exists ($I$ is the identity matrix), in that case multiplying the above equation from the left by $M^{-1}$ gives $a=b=c=0$ which is clearly not possible as $a,b,c$ are pairwise distinct, hence $\det(M) = 0$ must be true. Then $$0 = \det(M) = 1(1 \cdot 1 -(-y)(z))-(-x)(y(1)-(-z)(-y))+x(y(z)-(-z)(1)) = 1+yz+xy-xyz+xyz+xz = xy+yz+zx+1$$Hence $$xy+yz+xz = -1$$Then notice that $$x^2+y^2+z^2 = (x+y+z)^2-2(xy+yz+zx) = (x+y+z)^2+2 \qquad (\clubsuit)$$ We have the following claim to finish now: $\textbf{Claim:}$ $x+y+z \neq 0$ $\textbf{Proof)}$ Assume for the sake of contradiction that $$0 = x+y+z = \frac{a}{b-c}+\frac{b}{c-a}+\frac{a}{a-b} = \frac{a(c-a)(a-b)+b(b-c)(a-b)+c(b-c)(c-a)}{(b-c)(c-a)(a-b)}$$Hence the numerator which is $$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b = a^3+b^3+c^3+3abc \geq a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$$by Schur's inequality with equality only if $(a-b)(b-c)(c-a) = 0$ which is necessarily not the case, a contradiction to the fact that $x+y+z=0$. $\blacksquare$ Now, using this claim and $(\clubsuit)$, $$\frac{a^2}{(b-c)^2}+\frac{b^2}{(c-a)^2}+\frac{c^2}{(a-b)^2} = x^2+y^2+z^2 = (x+y+z)^2+2 > 2$$as desired. $\blacksquare$ $\blacksquare$