hello, with $b=a+u+v,c=a+u$ we get $ \left( 2\,{u}^{4}+4\,{u}^{3}v+6\,{u}^{2}{v}^{2}+4\,u{v}^{3}+2\,{v}^{4 } \right) {a}^{4}+ \left( 4\,{u}^{4}v+16\,{u}^{3}{v}^{2}+24\,{u}^{2}{v }^{3}+20\,u{v}^{4}+8\,{v}^{5} \right) {a}^{3}+ \left( 6\,{u}^{4}{v}^{2 }+24\,{u}^{3}{v}^{3}+42\,{u}^{2}{v}^{4}+36\,u{v}^{5}+12\,{v}^{6} \right) {a}^{2}+ \left( 4\,{u}^{4}{v}^{3}+20\,{u}^{3}{v}^{4}+36\,{u}^ {2}{v}^{5}+28\,u{v}^{6}+8\,{v}^{7} \right) a+2\,{u}^{4}{v}^{4}+8\,{u}^ {3}{v}^{5}+12\,{u}^{2}{v}^{6}+8\,u{v}^{7}+2\,{v}^{8}-3\,{u}^{4}{v}^{2} -6\,{u}^{3}{v}^{3}-3\,{u}^{2}{v}^{4}+2\,{u}^{4}+4\,{u}^{3}v+6\,{u}^{2} {v}^{2}+4\,u{v}^{3}+2\,{v}^{4} \geq0$ Sonnhard.

hello, for a) we get the value $1$. Sonnhard.

hello, for b) we get $-1$. Sonnhard.

gobathegreat wrote: Let $a$ ,$b$ and $c$ be distinct real numbers. $a)$ Determine value of $ \frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b} $ $b)$ Determine value of $ \frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b} $ $c)$ Prove the following inequality $ \frac{1+a^2b^2 }{(a-b)^2} + \frac{1+b^2c^2 }{(b-c)^2} + \frac{1+c^2a^2 }{(c-a)^2} \geq \frac{3}{2} $ By some simple calculations, we have \[ \frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b}=1 ,\] and \[ \frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b}=-1 .\] Now, rewrite the inequality in the form \[ \dfrac{(1+ab)^2+(1-ab)^2}{(a-b)^2}+\dfrac{(1+bc)^2+(1-bc)^2}{(b-c)^2}+\dfrac{(1+ca)^2+(1-ca)^2}{(c-a)^2} \geq 3.\] Notice that for all real numbers $x,y,z,$ we have $ x^2+y^2+z^2 \geq xy+yz+zx$ and $ x^2+y^2+z^2 \geq -2(xy+yz+zx).$ Therefore \[ \sum_{cyc}\dfrac{(1+ab)^2}{(a-b)^2} \geq \sum_{cyc}\frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} =1,\] and \[ \sum_{cyc}\dfrac{(1-ab)^2}{(a-b)^2} \geq -2\cdot \left[\sum_{cyc}\frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c}\right]=2.\] Hence, the proof is completed.

quykhtn-qa1 wrote: gobathegreat wrote: Let $a$ ,$b$ and $c$ be distinct real numbers. $a)$ Determine value of $ \frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b} $ $b)$ Determine value of $ \frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b} $ $c)$ Prove the following inequality $ \frac{1+a^2b^2 }{(a-b)^2} + \frac{1+b^2c^2 }{(b-c)^2} + \frac{1+c^2a^2 }{(c-a)^2} \geq \frac{3}{2} $ By some simple calculations, we have \[ \frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b}=1 ,\] and \[ \frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b}=-1 .\] Now, rewrite the inequality in the form \[ \dfrac{(1+ab)^2+(1-ab)^2}{(a-b)^2}+\dfrac{(1+bc)^2+(1-bc)^2}{(b-c)^2}+\dfrac{(1+ca)^2+(1-ca)^2}{(c-a)^2} \geq 3.\] Notice that for all real numbers $x,y,z,$ we have $ x^2+y^2+z^2 \geq xy+yz+zx$ and $ x^2+y^2+z^2 \geq -2(xy+yz+zx).$ Therefore \[ \sum_{cyc}\dfrac{(1+ab)^2}{(a-b)^2} \geq \sum_{cyc}\frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} =1,\] and \[ \sum_{cyc}\dfrac{(1-ab)^2}{(a-b)^2} \geq -2\cdot \left[\sum_{cyc}\frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c}\right]=2.\] Hence, the proof is completed. I am sorry I edited post When does equality holds?

Dr Sonnhard Graubner wrote: hello, with $b=a+u+v,c=a+u$ we get $ \left( 2\,{u}^{4}+4\,{u}^{3}v+6\,{u}^{2}{v}^{2}+4\,u{v}^{3}+2\,{v}^{4 } \right) {a}^{4}+ \left( 4\,{u}^{4}v+16\,{u}^{3}{v}^{2}+24\,{u}^{2}{v }^{3}+20\,u{v}^{4}+8\,{v}^{5} \right) {a}^{3}+ \left( 6\,{u}^{4}{v}^{2 }+24\,{u}^{3}{v}^{3}+42\,{u}^{2}{v}^{4}+36\,u{v}^{5}+12\,{v}^{6} \right) {a}^{2}+ \left( 4\,{u}^{4}{v}^{3}+20\,{u}^{3}{v}^{4}+36\,{u}^ {2}{v}^{5}+28\,u{v}^{6}+8\,{v}^{7} \right) a+2\,{u}^{4}{v}^{4}+8\,{u}^ {3}{v}^{5}+12\,{u}^{2}{v}^{6}+8\,u{v}^{7}+2\,{v}^{8}-3\,{u}^{4}{v}^{2} -6\,{u}^{3}{v}^{3}-3\,{u}^{2}{v}^{4}+2\,{u}^{4}+4\,{u}^{3}v+6\,{u}^{2} {v}^{2}+4\,u{v}^{3}+2\,{v}^{4} \geq0$ Sonnhard. Why it's true?

hello, we have $\left( 2\,{u}^{4}-3\,{u}^{2}+2 \right) {v}^{4}+ \left( 8\,{u}^{5}-6\, {u}^{3}+4\,u \right) {v}^{3}+ \left( 12\,{u}^{6}-3\,{u}^{4}+6\,{u}^{2} \right) {v}^{2}+ \left( 8\,{u}^{7}+4\,{u}^{3} \right) v+2\,{u}^{8}+2 \,{u}^{4} \geq 0$ since $2u^4-3u^2+2 \geq 0$ $8u^5-6u^3+4u\geq 0$ $12u^6-3u^4+6u^2\geq 0$ Sonnhard.

When $\begin{cases}\frac{1+bc }{b-c} = \frac{1+ca }{c-a} =\frac{1+ab}{a-b}\\ \frac{1-bc }{b-c} =\frac{1-ca }{c-a}=\frac{1-ab}{a-b} \end{cases}$ , the eqaulity holds.

gobathegreat wrote: Let $a$ ,$b$ and $c$ be distinct real numbers. $a)$ Determine value of $ \frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b} $ $b)$ Determine value of $ \frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b} $ $c)$ Prove the following ineqaulity $ \frac{1+a^2b^2 }{(a-b)^2} + \frac{1+b^2c^2 }{(b-c)^2} + \frac{1+c^2a^2 }{(c-a)^2} \geq \frac{3}{2} $ When does eqaulity holds? Generalization Let $a$ ,$b$ and $c$ be distinct real numbers. $a)$ Prove that $\frac{1+\lambda ab}{a-b}\cdot \frac{1+\lambda bc}{b-c}+\frac{1+\lambda bc}{b-c}\cdot \frac{1+\lambda ca}{c-a}+\frac{1+\lambda ca}{c-a}\cdot \frac{1+\lambda ab}{a-b}=\lambda ,\forall \lambda \in \mathbb{R}.$ $b)$ Prove the following ineqaulity $\frac{1+~{{\lambda }^{2}}{{a}^{2}}{{b}^{2}}}{{{(a-b)}^{2}}}+\frac{1+{{\lambda }^{2}}{{b}^{2}}{{c}^{2}}}{{{(b-c)}^{2}}}+\frac{1+{{\lambda }^{2}}{{c}^{2}}{{a}^{2}}}{{{(c-a)}^{2}}}\ge \frac{3\lambda }{2},\forall \lambda >0.$When does eqaulity holds?

Dr Sonnhard Graubner, we need to prove the starting inequality for reals variables.

ionbursuc wrote: gobathegreat wrote: Let $a$ ,$b$ and $c$ be distinct real numbers. $a)$ Determine value of $ \frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b} $ $b)$ Determine value of $ \frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b} $ $c)$ Prove the following ineqaulity $ \frac{1+a^2b^2 }{(a-b)^2} + \frac{1+b^2c^2 }{(b-c)^2} + \frac{1+c^2a^2 }{(c-a)^2} \geq \frac{3}{2} $ When does eqaulity holds? Generalization Let $a$ ,$b$ and $c$ be distinct real numbers. $a)$ Prove that $\frac{1+\lambda ab}{a-b}\cdot \frac{1+\lambda bc}{b-c}+\frac{1+\lambda bc}{b-c}\cdot \frac{1+\lambda ca}{c-a}+\frac{1+\lambda ca}{c-a}\cdot \frac{1+\lambda ab}{a-b}=\lambda ,\forall \lambda \in \mathbb{R}.$ $b)$ Prove the following ineqaulity $\frac{1+~{{\lambda }^{2}}{{a}^{2}}{{b}^{2}}}{{{(a-b)}^{2}}}+\frac{1+{{\lambda }^{2}}{{b}^{2}}{{c}^{2}}}{{{(b-c)}^{2}}}+\frac{1+{{\lambda }^{2}}{{c}^{2}}{{a}^{2}}}{{{(c-a)}^{2}}}\ge \frac{3\lambda }{2},\forall \lambda >0.$When does eqaulity holds? $a)$ Determine value of $ \frac{1+ab }{a-b} \cdot \frac{1+bc }{b-c} + \frac{1+bc }{b-c} \cdot \frac{1+ca }{c-a} + \frac{1+ca }{c-a} \cdot \frac{1+ab}{a-b} $ $b)$ Determine value of $ \frac{1-ab }{a-b} \cdot \frac{1-bc }{b-c} + \frac{1-bc }{b-c} \cdot \frac{1-ca }{c-a} + \frac{1-ca }{c-a} \cdot \frac{1-ab}{a-b} $ $c)$ Prove the following ineqaulity $ \frac{1+a^2b^2 }{(a-b)^2} + \frac{1+b^2c^2 }{(b-c)^2} + \frac{1+c^2a^2 }{(c-a)^2} \geq \frac{3}{2} $ \[\Longrightarrow\] Let $a$ ,$b$ and $c$ be distinct real numbers. $a)$ Prove that $\frac{1+\lambda ab}{a-b}\cdot \frac{1+\lambda bc}{b-c}+\frac{1+\lambda bc}{b-c}\cdot \frac{1+\lambda ca}{c-a}+\frac{1+\lambda ca}{c-a}\cdot \frac{1+\lambda ab}{a-b}=\lambda ,\forall \lambda \in \mathbb{R}.$ $b)$ Prove the following ineqaulity $\frac{1+~{{\lambda }^{2}}{{a}^{2}}{{b}^{2}}}{{{(a-b)}^{2}}}+\frac{1+{{\lambda }^{2}}{{b}^{2}}{{c}^{2}}}{{{(b-c)}^{2}}}+\frac{1+{{\lambda }^{2}}{{c}^{2}}{{a}^{2}}}{{{(c-a)}^{2}}}\ge \frac{3\lambda }{2},\forall \lambda >0.$

sqing wrote: When $\begin{cases}\frac{1+bc }{b-c} = \frac{1+ca }{c-a} =\frac{1+ab}{a-b}\\ \frac{1-bc }{b-c} =\frac{1-ca }{c-a}=\frac{1-ab}{a-b} \end{cases}$ , the eqaulity holds. When $\begin{cases}\frac{1+bc }{b-c} = \frac{1+ca }{c-a} =\frac{1+ab}{a-b}\\ \frac{1-bc }{b-c} +\frac{1-ca }{c-a}+\frac{1-ab}{a-b}=0 \end{cases}$ , the eqaulity holds.

Refinement $\frac{1+{{a}^{2}}{{b}^{2}}}{{{(a-b)}^{2}}}+\frac{1+{{b}^{2}}{{c}^{2}}}{{{(b-c)}^{2}}}+\frac{1+{{c}^{2}}{{a}^{2}}}{{{(c-a)}^{2}}}=\frac{1}{2}\left( 3+\frac{1}{2}\sum\limits_{{}}{{{\left( \frac{1+ab}{a-b}-\frac{1+bc}{b-c} \right)}^{2}}+{{\left( \sum\limits_{{}}{\frac{1-ab}{a-b}} \right)}^{2}}} \right)\ge $ $\frac{1}{2}\left( 3+{{\left( \sum\limits_{{}}{\frac{1-ab}{a-b}} \right)}^{2}} \right)$

Generalization and refinement $\frac{1+{{\lambda }^{2}}{{a}^{2}}{{b}^{2}}}{{{(a-b)}^{2}}}+\frac{1+{{\lambda }^{2}}{{b}^{2}}{{c}^{2}}}{{{(b-c)}^{2}}}+\frac{1+{{\lambda }^{2}}{{c}^{2}}{{a}^{2}}}{{{(c-a)}^{2}}}=\frac{1}{2}\left( 3\lambda +\frac{1}{2}\sum\limits_{{}}{{{\left( \frac{1+\lambda ab}{a-b}-\frac{1+\lambda bc}{b-c} \right)}^{2}}+{{\left( \sum\limits_{{}}{\frac{1-\lambda ab}{a-b}} \right)}^{2}}} \right)\ge $$\frac{1}{2}\left( 3\lambda +{{\left( \sum\limits_{{}}{\frac{1-\lambda ab}{a-b}} \right)}^{2}} \right)$