Advertisements
Advertisements
प्रश्न
Prove that a2 + b2 + c2 − ab − bc − ca is always non-negative for all values of a, b and c
Advertisements
उत्तर
We have
`a^2 + b^2 + c^2 - ab - bc - ca`
`= 2/2[a^2 + b^2 + c^2 - ab - bc - ca]` [Mulitply and divide by 2]
`= 1/2 [2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca]`
`= 1/2 [a^2 + a^2 + b^2 + b^2 + c^2 - 2ab - 2bc - 2ac]`
`= 1/2[(a^2 + b^2 - 2ab) + (a^2 + c^2 - 2ac) + (b^2 + c^2 - 2bc)]`
`= 1/2 [(a - b)^2 + (b - c)^2 + (c - a)^2]` `[∵ (a - b)^2 = a^2 + b^2 - 2ab]`
`= ((a - b)^2 + (b -c)^2 + (c - a)^2)/2 >= 0`
`∴ a^2 + b^2 + c^2 - ab - bc -ca >= 0`
hence `a^2 + b^2 - ab - bc - ca > 0`
Hence `a^2 + b^2 + c^2 - ab - bc - ca` is always non-negative for all values of a, b and c.
APPEARS IN
संबंधित प्रश्न
Use suitable identity to find the following product:
(x + 4) (x + 10)
Use suitable identity to find the following product:
`(y^2+3/2)(y^2-3/2)`
Factorise the following using appropriate identity:
`x^2 - y^2/100`
Evaluate the following using identities:
(1.5x2 − 0.3y2) (1.5x2 + 0.3y2)
Simplify the following products:
`(x^3 - 3x^2 - x)(x^2 - 3x + 1)`
Write in the expanded form: `(x/y + y/z + z/x)^2`
Simplify (a + b + c)2 + (a - b + c)2 + (a + b - c)2
Find the value of 4x2 + y2 + 25z2 + 4xy − 10yz − 20zx when x = 4, y = 3 and z = 2.
If \[x^2 + \frac{1}{x^2}\], find the value of \[x^3 - \frac{1}{x^3}\]
Evaluate of the following:
933 − 1073
If \[x + \frac{1}{x}\] 4, then \[x^4 + \frac{1}{x^4} =\]
If a + b + c = 9 and ab + bc + ca =23, then a3 + b3 + c3 − 3abc =
If a + b = 7 and ab = 10; find a - b.
Use the direct method to evaluate :
(3b−1) (3b+1)
Use the direct method to evaluate :
(ab+x2) (ab−x2)
Evaluate: (2 − z) (15 − z)
Evaluate: (5xy − 7) (7xy + 9)
If a2 - 3a - 1 = 0 and a ≠ 0, find : `"a" - (1)/"a"`
Simplify:
(x + 2y + 3z)(x2 + 4y2 + 9z2 - 2xy - 6yz - 3zx)
Simplify:
(3x + 5y + 2z)(3x - 5y + 2z)
