Question

If \[\lambda x+( 1/x )-1 \] is always non negative for x>0 and \[\lambda > 0\] ,then the least value of \[\lambda \] is ?

Answer 1

here is my idea
start with
\[\lambda x+\frac{1}{x}-1>0\] since \(x>0\) can multiply by \(x\) and get \[\lambda x^2+x-1>0\]

Answer 2

-x would be there instead of -1 i think

Answer 3

parabola that opens up, make sure the vertex is above the \(x\) axis
thust a thought though

Answer 4

since you multiplied that with x

Answer 5

oops

Answer 6

\(\lambda x^2+x-1\) looks better

Answer 7

\[\lambda x^2+x-x>0\]

Answer 8

man i still can't write correctly
let me try one more time
\[\lambda x^2-x+1>0\] how about that one??

Answer 9

\[\lambda x^2>0\]

Answer 10

@satellite73 Lol... Yes that is correct.

Answer 11

i'll use y for (lambda)
its given
yx + 1/x - 1 >= 0
y>= 1/x - 1/x^2
or
y>= (x-1)/x^2
we need to find least value of y or in other words max value of (x-1)/x^2
if f(x) = 1/x - 1/x^2
then f'(x) = -1/x^2 + 2/x^3
and f''(x) = 2/x^3 - 6/x^4

for f'(x)= 0, we have x=2
and f''(2) < 0
thus x=2 will be a maxima for (x-1)/x^2

thus min value of y will be 1/4 ..

Answer 12

\[\large{\lambda x+\frac{1}{x}-1>0}\]
\[\large{x*(\lambda x+\frac{1}{x}-1)>0}\]
\[\large{\lambda x^x+x-x>0}\]

Answer 13

@shubhamsrg 's calc answer makes good sense. Visually, it checks out too.

Answer 14

@shubhamsrg How is the least value of y the maximum value of the right side expression? Shouldn't (x-1)/x^2 also be minimum?
I get the rest of the answer though.

Answer 15

@mathteacher1729 do you have any idea why the minimum value of y is the maximum value of (x-1)/x^2 ?

Answer 16

lets consider the graphs ,,i dont want to figure out what graph RHS will have,,so i take any arbitrary one..we plot x vs lambda

the shaded reigon is where lambda>= RHS
so now you must be getting why RHS needs to be max for LHS to be min..

sorry had got DC that time..