Question

If ax+b/x >=c for all +ve values of a,b and c then

Answer 1

\[A)~ab \ge \frac{c^2}{4}\]
B)ab<c^2/4
C)bc>a^2/4
D)ac>b^2/4

Answer 2

I used AM>GM and got B but it is wrong somehow,m i commiting a silly mistake?

Answer 3

\[\large \frac{ax+\frac{b}{x}}{2}>\sqrt{ab}=>\frac{c^2}{4}>ab\]

Answer 4

$$ax+\frac{b}x\ge c\\ax^2+b\ge cx\\ax^2-cx+b\ge 0$$This tells us our discriminant is nonpositive:$$c^2-4ab\le0\\c^2\le4ab\\\frac{c^2}4\le ab$$

Answer 5

for \(x<0\) that should read $$ax^2+b\le cx\\ax^2-cx+b\le 0$$but the rest still works; this inequality tells us we have at most 1 distinct real root

Answer 6

"This tells us our discriminant is nonpositive:" how?

Answer 7

@DLS $$ax^2+bx+c=0\\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$if our function has at most 1 distinct root, we either have \(b^2-4ac=0\) or we have no real roots i.e. \(b^2-4ac<0\)

Answer 8

how do we know we have no real roots?

Answer 9

@DLS nobody said we have none, we just know we have at most 1 because it never crosses the \(x\)-axis, only touches...