Three Real numbers

Geometry Level 2

If a , b , c a, b, c are positive real numbers such that no two of them are equal

Then the Equation a ( a − b ) ( a − c ) + b ( b − c ) ( b − a ) + c ( c − a ) ( c − b ) a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b) is always

Positive Zero Can't Say Anything Negative

Rishabh Deep Singh by Rishabh Deep Singh , 23, India

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1 solution

L e t a > b > c Let \quad a > b > c a ( a − b ) ( a − c ) − b ( b − c ) ( a − b ) + c ( c − a ) ( c − b ) a(a - b)(a - c) - b(b - c)(a - b) + c(c - a)(c - b) ( a − b ) [ a ( a − c ) − b ( b − c ) ] + c ( c − a ) ( c − b ) (a - b)[a(a - c) - b(b - c)] + c(c - a)(c - b) ( a − b ) [ a 2 − a c − b 2 + b c ] + c ( c − a ) ( c − b ) (a - b)[a2 - ac - b2 + bc] + c(c - a) (c - b) ( a − b ) [ a 2 − b 2 − ( a c − b c ) ] + c ( c − a ) ( c − b ) (a - b)[a2 - b2 - (ac - bc)] + c (c - a) (c - b) ( a − b ) [ ( a − b ) ( a + b ) − c ( a − b ) ] + c ( c − a ) ( c − b ) (a - b)[(a - b) (a + b) - c (a - b)] + c (c - a) (c - b) ( a − b ) 2 ( a + b − c ) ⏟ X + c ( c − a ) ( c − b ) ⏟ Y \underbrace{(a-b)^2(a+b-c)}_{X}+\underbrace{c(c-a)(c-b)}_{Y} X > 0 a n d Y > 0 X >0 \quad and\quad Y > 0 ∴ X + Y > 0 \therefore X+Y > 0

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