cho a,b,c >0 tm ab +bc+ca =1 cmr a/a ²+1+b/b ²+1+c/c ²+1=2/ √(a ²+1)(b ²+1)(c ²+1) 13/07/2021 Bởi Natalia cho a,b,c >0 tm ab +bc+ca =1 cmr a/a ²+1+b/b ²+1+c/c ²+1=2/ √(a ²+1)(b ²+1)(c ²+1)
$\begin{array}{l} ab + bc + ca = 1\\ \Rightarrow \left\{ \begin{array}{l} {a^2} + 1 = {a^2} + ab + bc + ca = \left( {a + b} \right)\left( {a + c} \right)\\ {b^2} + 1 = {b^2} + ab + bc + ca = \left( {b + a} \right)\left( {b + c} \right)\\ {c^2} + 1 = {c^2} + ab + bc + ca = \left( {c + a} \right)\left( {b + c} \right) \end{array} \right.\\ \dfrac{a}{{{a^2} + 1}} + \dfrac{b}{{{b^2} + 1}} + \dfrac{c}{{{c^2} + 1}}\\ = \dfrac{a}{{\left( {a + b} \right)\left( {a + c} \right)}} + \dfrac{b}{{\left( {b + c} \right)\left( {b + a} \right)}} + \dfrac{c}{{\left( {c + a} \right)\left( {c + b} \right)}}\\ = \dfrac{{a\left( {b + c} \right) + b\left( {a + c} \right) + c\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} = \dfrac{{2\left( {ab + bc + ca} \right)}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}}\\ = \dfrac{2}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} = \dfrac{2}{{\sqrt {{{\left( {a + b} \right)}^2}{{\left( {b + c} \right)}^2}{{\left( {c + a} \right)}^2}} }}\\ = \dfrac{2}{{\sqrt {\left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\left( {{c^2} + 1} \right)} }} \end{array}$ Bình luận
$\begin{array}{l} ab + bc + ca = 1\\ \Rightarrow \left\{ \begin{array}{l} {a^2} + 1 = {a^2} + ab + bc + ca = \left( {a + b} \right)\left( {a + c} \right)\\ {b^2} + 1 = {b^2} + ab + bc + ca = \left( {b + a} \right)\left( {b + c} \right)\\ {c^2} + 1 = {c^2} + ab + bc + ca = \left( {c + a} \right)\left( {b + c} \right) \end{array} \right.\\ \dfrac{a}{{{a^2} + 1}} + \dfrac{b}{{{b^2} + 1}} + \dfrac{c}{{{c^2} + 1}}\\ = \dfrac{a}{{\left( {a + b} \right)\left( {a + c} \right)}} + \dfrac{b}{{\left( {b + c} \right)\left( {b + a} \right)}} + \dfrac{c}{{\left( {c + a} \right)\left( {c + b} \right)}}\\ = \dfrac{{a\left( {b + c} \right) + b\left( {a + c} \right) + c\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} = \dfrac{{2\left( {ab + bc + ca} \right)}}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}}\\ = \dfrac{2}{{\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)}} = \dfrac{2}{{\sqrt {{{\left( {a + b} \right)}^2}{{\left( {b + c} \right)}^2}{{\left( {c + a} \right)}^2}} }}\\ = \dfrac{2}{{\sqrt {\left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\left( {{c^2} + 1} \right)} }} \end{array}$