Question

prove that (a+b+c)^2>3(ab+bc+ca)

Answer 1

$\large\underline{\sf{Solution-}}$

Without loss of inequality, assume that

$\rm :\longmapsto\:a \: = \: b \: = \: c$

Then,

$\rm :\longmapsto\: {(a + a + a)}^{2} { > } \: 3({a}^{2} + {a}^{2} + {a}^{2})$

$\rm :\longmapsto\: {(3a)}^{2} { > } \: 9{a}^{2}$

$\rm :\longmapsto\: {9a}^{2} { > } \: 9{a}^{2}$

Which is not true.

$\bf\implies \:\:a \: \ne \: b \: \ne \: c$

We know that

$\rm :\longmapsto\: {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$

So, this can be rewritten as

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = {(a + b + c)}^{2} - 2(ab + bc + ca) - - - (1)$

Now,

Consider,

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca$

can be rewritten as

$\rm \:  =  \:  \: \dfrac{1}{2}(2{a}^{2} + 2 {b}^{2} +2 {c}^{2} - 2ab -2 bc - 2ca)$

$\rm \: =\dfrac{1}{2}({a}^{2} + {a}^{2} + {b}^{2} + {b}^{2} +{c}^{2} + {c}^{2} - 2ab -2 bc - 2ca)$

$\rm \:  = \dfrac{1}{2}\bigg(( {a}^{2} + {b}^{2} - 2ab) + ( {b}^{2} + {c}^{2} - 2bc) + ( {c}^{2} + {a}^{2} - 2ac)\bigg)$

$\rm \:  = \dfrac{1}{2}\bigg( {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}\bigg)$

As sum of squares can never be negative, and a, b, c are distinct,

$\bf\implies \: {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca > 0$

$\bf\implies \: {a}^{2} + {b}^{2} + {c}^{2} > ab + bc + ca$

Now, On substituting the value from equation (1), we get

$\bf\implies \: {(a + b + c)}^{2} - 2(ab + bc + ca)> ab + bc + ca$

$\bf\implies \: {(a + b + c)}^{2} > 2(ab + bc + ca) + ab + bc + ca$

$\bf\implies \: {(a + b + c)}^{2} > 3(ab + bc + ca)$

Hence, Proved