Prove that if 1<a<2 , that 2<a+1/a<3? please , i really need your help!!!

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2 Answers
Apr 16, 2018

See explanation

Explanation:

Let a=p/q where p and q are positive integers.

1ltp/q therefore qltp. p/qlt2 therefore plt2q. Therefore qltplt2q.

a+1/a=p/q+q/p=(pp)/(qp)+(qq)/(pq)=(p^2+q^2)/(pq)=(p^2+2pq+q^2-2pq)/(pq)=(p+q)^2/(pq)-(2pq)/(pq)=(p+q)^2/(pq)-2

(q+q)^2/(qq)lt(p+q)^2/(pq)lt(2q+q)^2/(2qq)*
(2q)^2/q^2lt(p+q)^2/(pq)lt(3q)^2/(2q^2)
(4q^2)/q^2lt(p+q)^2/(pq)lt(9q^2)/(2q^2)
4lt(p+q)^2/(pq)lt9/2
4-2lt(p+q)^2/(pq)-2lt9/2-2
2lt(p+q)^2/(pq)-2lt5/2
2lta+1/alt5/2

5/2lt6/2
5/2lt3
2lta+1/alt3

~~More advanced topics ahead~~

*This assumes that as p increases, (p+q)^2/(pq) increases. This can be verified intuitively, by looking at the graph of y=(x+q)^2/(xq) on x in(q,2q) for various positive values of q, or by the calculus process below.

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del/(delp)[(p+q)^2/(pq)]=1/qdel/(delp)[(p+q)^2/p]=1/q(pdel/(delp)[(p+q)^2]-(p+q)^2del/(delp)[p])/p^2=1/q(p[2(p+q)]-(p+q)^2[1])/p^2=1/q(2p(p+q)-(p+q)^2)/p^2=((2p^2+2pq)-(p^2+2pq+q^2))/(p^2q)=(p^2-q^2)/(p^2q).

On p in(q,2q):
Since pgtqgt0, p^2gtq^2 thus p^2-q^2gt0.
Since q>0, p^2qgt0
Since p^2-q^2gt0 and p^2qgt0, (p^2-q^2)/(p^2q)gt0
Since del/(delp)[(p+q)^2/(pq)]=(p^2-q^2)/(p^2q) and (p^2-q^2)/(p^2q)gt0, del/(delp)[(p+q)^2/(pq)]gt0

Therefore (p+q)^2/(pq) is increasing for constant q and qltplt2q because del/(delp)[(p+q)^2/(pq)] is positive.

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Apr 16, 2018

In description

Explanation:

Here constraint (1):

1 < a < 2

Constraint(2):
By reciprocal theorem,

1/1 > 1/a > 1/2
1 > a> 1/2

In constraint 1 add 1 on both sides,
1+1 < a+1 < 2+1
2 < a+1 < 3

color (red)(a+1 <3)

In same constraint add 1/2

(1+1/2)<(a+1/2)<(2+1/2)

Again note that, 2 <2+1/2
So a+1/2 must be less than 2
color (red)(a+1/2)<2

Hence In constraint 2,
1 > a> 1/2
Add a on both sides,

1+a > a+1/a > 1/2+a
3 > a+1/a >2
2 < a+1/a <3

We did it so because a+1<3
So a+1/a must be less than 3.

Again a+1/2<2 but in this constraint a+1/a > a+1/2
So, a+1/a must be greater than 2.
Hence, 1> 1/a > 1\2

By adding a on both sides,
1+a > a+1/a > a+1/2
3 > a+1/a > 2
2 < a+1/a < 3 proved