mavropnevma wrote: For all triplets $a,b,c$ of (pairwise) distinct real numbers, prove the inequality $$ \left | \dfrac {a} {b-c} \right | + \left | \dfrac {b} {c-a} \right | + \left | \dfrac {c} {a-b} \right | \geq 2$$ and determine all cases of equality. Let $\frac{a}{b-c}=x$, $\frac{b}{c-a}=y$ and $\frac{c}{a-b}=z$. Hence, $xy+xz+yz=-1$ and $\left(|x|+|y|+|z|\right)^2=(x+y+z)^2+2+2\sum_{cyc}|xy|\geq4$. The equality occurs for $x+y+z=0$ and $\sum_{cyc}|xy|=1$, which hods.

For all triplets $a,b,c,d$ of (pairwise) distinct real numbers, prove the inequality \[ \left | \dfrac {a} {b-c} \right | + \left | \dfrac {b} {c-d} \right | + \left | \dfrac {c} {d-a} \right | + \left | \dfrac {d} {a-b} \right | \geq 2.\]

sqing wrote: For all triplets $a,b,c,d$ of (pairwise) distinct real numbers, prove the inequality \[ \left | \dfrac {a} {b-c} \right | + \left | \dfrac {b} {c-d} \right | + \left | \dfrac {c} {d-a} \right | + \left | \dfrac {d} {a-b} \right | \geq 2.\] See here: http://www.artofproblemsolving.com/community/c6h614833

it s not dfficult to show that The equality occurs iff one of the numbers a,b or c is zero and the others are opposite numbers

Vasc inequality Let $a,b,c$ be real numbers such that $abc\neq 0$ . Prove the inequality $$ \left | \dfrac{b+c} {a} \right | + \left | \dfrac {c+a} {b}\right | + \left | \dfrac {a+b} {c} \right | \geq 2.$$

sqing wrote: Vasc inequality Let $a,b,c$ be real numbers such that $abc\neq 0$ . Prove the inequality $$ \left | \dfrac{b+c} {a} \right | + \left | \dfrac {c+a} {b}\right | + \left | \dfrac {a+b} {c} \right | \geq 2.$$ Among the three real numbers $a,b$ and $c$ two have the same sign and if $(a,b,c)$ satisfies the inequality then $(-a,-b-c)$ satisfies it too. WLOG suppose that $a, b >0$ and $ a \leq b$. If $c>0$ then the left hand side is greater or equal than 6 by direct application of AM-GM If $c<0$ let $c'=-c>0$ then we have to prove for positive real numbers $a,b,c'$ that : $$ \dfrac{|b-c'|} {a} + \dfrac {|a-c'|} {b}+\dfrac {a+b} {c'} \geq 2.$$ -If $c' \in [a,b]$ then $ \dfrac{|b-c'|} {a} + \dfrac {|a-c'|} {b}+\dfrac {a+b} {c'}= \dfrac{b-c'} {a} + \dfrac {c'-a} {b}+\dfrac {a+b} {c'}=c'\left(\frac{1}{b}-\frac{1}{a}\right)+\frac{a+b}{c'}+\frac{b}{a}-\frac{a}{b}$ is a decreasing function of $c'$ so it is minimal when $c'=b$ and equal to $2$ in this case. -If $c'\leq a$ then $ \dfrac{|b-c'|} {a} + \dfrac {|a-c'|} {b}+\dfrac {a+b} {c'}= \dfrac{b-c'} {a} + \dfrac {a-c'} {b}+\dfrac {a+b} {c'}=-c'\left(\frac{1}{b}+\frac{1}{a}\right)+\frac{a+b}{c'}+\frac{b}{a}+\frac{a}{b}$ is decreasing function of $c'$ so it is minimal when $c'=a$ and equal to $2\frac{b}{a}$ in this case and which it is greater than $2$. -If $c' \geq b$ then $ \dfrac{|b-c'|} {a} + \dfrac {|a-c'|} {b}+\dfrac {a+b} {c'}= \dfrac{c'-b} {a} + \dfrac {c'-a} {b}+\dfrac {a+b} {c'}=(a+b)\left(\frac{c'}{ab}+\frac{1}{c'}\right)-\left(\frac{b}{a}+\frac{a}{b}\right)$ is increasing function of $c'$ on $(b, +\infty)$ (because $c'^2>ab$) and minimal when $c'=b$ and equal to $2$ in this case.

sqing wrote: Vasc inequality Let $a,b,c$ be real numbers such that $abc\neq 0$ . Prove the inequality $$ \left | \dfrac{b+c} {a} \right | + \left | \dfrac {c+a} {b}\right | + \left | \dfrac {a+b} {c} \right | \geq 2.$$ Thank tchebytchev. Proof of Vasc: Let $|a|=max\{|a|,|b|,|c|\}$, hence $ \left | \dfrac{b+c} {a} \right | + \left | \dfrac {c+a} {b}\right | + \left | \dfrac {a+b} {c} \right | \geq\dfrac{ \left | b+c\right |} { \left | a\right |} + \dfrac { \left | c+a\right |} { \left | a\right |}+ \dfrac { \left | a+b\right |} { \left | a\right |} \geq \dfrac{\left | -b-c+c+a+a+b\right | } {\left | a \right | }=2$, the equality holds for $a=1,b=-1$ and $|c|\leq1$ (or any permutation).

sqing wrote: sqing wrote: Vasc inequality Let $a,b,c$ be real numbers such that $abc\neq 0$ . Prove the inequality $$ \left | \dfrac{b+c} {a} \right | + \left | \dfrac {c+a} {b}\right | + \left | \dfrac {a+b} {c} \right | \geq 2.$$ Thank tchebytchev. Proof of Vasc: Let $|a|=max\{|a|,|b|,|c|\}$, hence $ \left | \dfrac{b+c} {a} \right | + \left | \dfrac {c+a} {b}\right | + \left | \dfrac {a+b} {c} \right | \geq\dfrac{ \left | b+c\right |} { \left | a\right |} + \dfrac { \left | c+a\right |} { \left | a\right |}+ \dfrac { \left | a+b\right |} { \left | a\right |} \geq \dfrac{\left | -b-c+c+a+a+b\right | } {\left | a \right | }=2$, the equality holds for $a=1,b=-1$ and $|c|\leq1$ (or any permutation). Direct and elegant solution

sqing wrote: sqing wrote: Vasc inequality Let $a,b,c$ be real numbers such that $abc\neq 0$ . Prove the inequality $$ \left | \dfrac{b+c} {a} \right | + \left | \dfrac {c+a} {b}\right | + \left | \dfrac {a+b} {c} \right | \geq 2.$$ ... the equality holds for $a=1,b=-1$ and $|c|\leq1$ (or any permutation). the equality holds also for $ a=-b, |c|\le|a| $ or any permutation

Mavro....pnevma... You are no less than god.

Stars of Mathematics 2012, Seniors, Problem 3 mavropnevma wrote: For all triplets $a,b,c$ of (pairwise) distinct real numbers, prove the inequality $$ \left | \dfrac {a} {b-c} \right | + \left | \dfrac {b} {c-a} \right | + \left | \dfrac {c} {a-b} \right | \geq 2$$and determine all cases of equality. Prove that if we also impose $a,b,c$ positive, then all equality cases disappear, but the value $2$ remains the best constant possible. (Dan Schwarz) https://artofproblemsolving.com/community/c6h1061432p4596239: Stars of Mathematics 2012, Juniors, Problem 3 mavropnevma wrote: For all triplets $a,b,c$ of (pairwise) distinct real numbers, prove the inequality $$ \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | \geq 2$$and determine all cases of equality. Prove that if we also impose $a,b,c \geq 0$, then $$ \left | \dfrac {a+b} {a-b} \right | + \left | \dfrac {b+c} {b-c} \right | + \left | \dfrac {c+a} {c-a} \right | > 3,$$with the value $3$ being the best constant possible. (Dan Schwarz) Stars of Mathematics: Romania Stars of Mathematics (Romania)

sqing wrote: Vasc inequality Let $a,b,c$ be real numbers such that $abc\neq 0$ . Prove the inequality $$ \left | \dfrac{b+c} {a} \right | + \left | \dfrac {c+a} {b}\right | + \left | \dfrac {a+b} {c} \right | \geq 2.$$ Let $n\geqslant 3$ be an integer and $a_1,\ldots,a_n$ be nonzero real numbers, with sum $S{}$. Prove that\[\sum_{i=1}^n\left|\frac{S-a_i}{a_i}\right|\geqslant\frac{n-1}{n-2}.\]

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