341(a^4+b^4)x^2+4abcdx + (c^4+d^4) \ge 2a^2b^2 x^2 + 4abcdx + 2c^2d^2 \ \text{from AM-GM Ineq}
\ge 2 (abx + cd)^2 \ge 0
Note that
(i) Equality occurs iff a=b and c=d
(ii) There is a unique value of x for which the expression then becomes zero
q.e.d
Prove that the roots of the equation (a4+b4)x2+4abcdx+(c4+d4)=0 cannot be different if real.
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2 Answers
1finding the discriminant of the given quadratic equation
D= (4abcd)2 -4(a4+b4)(c4+d4)
D= 4{4a2b2c2d2 -(a4c4+a4d4+b4c4+b4d4)
D= -4{ a4c4 +b4d4 - 2a2b2c2d2 + a4d4 + b4c4 - 2 a2b2c2d2 }
D= -4 { (a2c2 -b2d2)2 + (a2d2-b2c2)2 }
thus ,
D<=0
for roots to be real
D>=0
thus for real roots
D=0 => roots are equal
Hence if roots are real they are equal
341