Prove that the minimum value of `((a+x)(b+x))/((c+x))a ,b > c ,x >-c` is `(sqrt(a-c)+sqrt(b-c))^2`

Show that the minimum value of (x+a)(x+b)//(x+c)dotw h e r ea > c ,b > c , is (sqrt(a-c)+sqrt(b-c))^2 for real values of x > -c

Vertices of a variable acute angled triangle A B C lies on a fixed circle. Also, a ,b ,ca n dA ,B ,C are lengths of sides and angles of triangle A B C , respectively. If x_1, x_2a n dx_3 are distances of orthocenter from A ,Ba n dC , respectively, then the maximum value of ((dx_1)/(d a)+(dx_2)/(d b)+(dx_3)/(d c)) is -sqrt(3) b. -3sqrt(3) c. sqrt(3) d. 3sqrt(3)

The minimum value of P=b c x+c a y+a b z , when x y z=a b c , is

If a^2x^4+b^2y^4=c^6, then the maximum value of x y is (a) (c^2)/(sqrt(a b)) (b) (c^3)/(a b) (c) (c^3)/(sqrt(2a b)) (d) (c^3)/(2a b)

If a ,b ,c are in A.P., then prove that the following are also in A.P. 1/(sqrt(b)+sqrt(c)),1/(sqrt(c)+sqrt(a)),1/(sqrt(a)+sqrt(b))

Let f(x)=a bsinx+bsqrt(1-a^2)cosx+c , where |a| >0 then (a) c-b,c+b (b) difference of maximum and minimum values of f(x) is 2b (c) f(x)=c if x= -cos^(-1)a (d) f(x)=c if x= cos^(-1)a

If y=((a-x)sqrt(a-x)-(b-x)sqrt(x-b))/((sqrt(a-x)+sqrt(x-b)) ,then (dy)/(dx) wherever it is defined is (a) (x+(a+b))/(sqrt((a-x)(x-b))) (b) (2x-a-b)/(2sqrt(a-x)sqrt(x-b)) (c) -((a+b))/(2sqrt((a-x)(x-b))) (d) (2x+(a+b))/(2sqrt((a-x)(x-b)))

If c!=0 and the equation p//(2x)=a//(x+c)+b//(x-c) has two equal roots, then p can be (sqrt(a)-sqrt(b))^2 b. (sqrt(a)+sqrt(b))^2 c. a+b d. a-b

Prove that x^((b-c)/(bc)) x^((c-a)/(ca)) x^((a-b)/(ab))=1

Prove that the greatest value of x y is c^3//sqrt(2a b)dot if a^2x^4+b^4y^4=c^6dot