Question

How to prove that this set of inequalities is satisfied for all integer x?
\[\sqrt{x^2+1}+\sqrt{x^2+2x}>2x+1\\
\sqrt{x^2+2}+\sqrt{x^2+2x-1}>2x+1\\
\vdots\\
\sqrt{x^2+x}+\sqrt{x^2+x+1}>2x+1\]

Answer 1

x=0 to -2, \( \sqrt{x^2+2x} \) does not have real value

Answer 2

I think one suppose to do it for x positive

Answer 3

I meant for positive integer x. My bad.

Answer 4

i dont get the law of formation

Answer 5

ok i get it

Answer 6

prove it's valid for 1
\[\sqrt{1+1}+\sqrt{1+2}>3\]
\[\sqrt{1+2}+\sqrt{1+2-1}>3\]

Answer 7

and so on

Answer 8

now supose it's valid for a K

Answer 9

and now we have to prove for K+1

Answer 10

What's u in your proof above?

Answer 11

k+1

Answer 12

i forget exp 2

Answer 13

\[\sqrt{(K+1)^{2}+1}+\sqrt{(K+1)^{2}+2.(K+1)}=\sqrt{u^{2}+1}+\sqrt{u^{2}+2.u}>2.u+1\]
and so on
so we prove it's valid for K+1,for induction it's valid for all intergers

Answer 14

Well, this isn't the most elegant method, but do this: square the first inequality and simplify. Then square again. The problem should reduce to a simple one.

Answer 15

Isn't there a magical inequality out there that can easily solve this problem?

Answer 16

There are a couple of standard tricks, but I don't see how they would apply to your problem. For instance, using Cauchy-Schwartz
\[ \sqrt{x^2 + 1} + \sqrt{x^2 + 2x} = 1 \cdot \sqrt{x^2 + 1} + 1 \cdot \sqrt{x^2 + 2x} \leq \sqrt{2} (x^2 + 1 + x^2 + 2x ) \]
but as you can see, this has the inequality going the 'wrong way'.

I'll think about it a little more.

Answer 17

Let
\[
f(x)= \sqrt{x^2+2x-y+1}+\sqrt{x^2+y}, \quad 0<y < x\\

g(x) = 2 x +1\\

f(x)^2 > g(x)^2 \\

2 \sqrt{x^2+2 x-y+1}
\sqrt{x^2+y}+2 x^2+2 x+1>4
x^2+4 x+1\\

2 \sqrt{x^2+2 x-y+1}
\sqrt{x^2+y}>2 x^2+2 x\\
\sqrt{x^2+2 x-y+1}
\sqrt{x^2+y}> x^2+2 x\\

(\sqrt{x^2+2 x-y+1})
\sqrt{x^2+y})^2> ( x^2+2 x)^2\\
x^4+2 x^3+x^2+2 x
y-y^2+y>x^4+2 x^3+x^2\\

2 x y-y^2+y>0 \\
\text { We need to show the last inequality }\\
2 x y -y^2 + y > 2 y^2 -y^2 + y= 2 y^2 + y >0



\]

Answer 18

\[\sqrt{x ^{2}+x} + \sqrt{x ^{2} + x +1} > 2x+1\]
square both the sides...
we get
\[x ^{2} + x +x ^{2} + x +1 + 2\sqrt{x ^{2}+x}\sqrt{x ^{2}+x+1} > 4x ^{2} + 4x +1\]
join like terms to get
\[2\sqrt{x ^{2}+x}\sqrt{x ^{2}+x+1} > 2x ^{2}+2x\]
strike off 2 from all the sides and square th sides again...
\[(x ^{2}+x)(x ^{2}+x+1)>(x ^{2} + x)^{2}\]
since for positive x
\[(x ^{2}+x)\] is positive
we can strike that term off from both sides to get
\[x ^{2} + x +1 > x ^{2} + x\]
and hence we get 1 > 0
which is a universal truth....so the given inequality would be satisfied for all positive integers...

Answer 19

Let k be a positive number (k<=x-1)\[(k+1)(2x-k)>0\]\[2kx+2x-k^2-k>0\]\[x^4+2x^3+x^2+2kx+2x-k^2-k>x^4+2x^3+x^2\]\[(x^2+1+k)(x^2+2x-k)>(x^2+x)^2\]\[2\sqrt{(x^2+1+k)(x^2+2x-k)}>2x^2+2x\]\[2x^2+2x+1+2\sqrt{(x^2+1+k)(x^2+2x-k)}>4x^2+4x+1\]\[x^2+1+k+x^2+2x-k+2\sqrt{(x^2+1+k)(x^2+2x-k)}>(2x+1)^2\]\[(\sqrt{x^2+1+k}+\sqrt{x^2+2x-k})^2>(2x+1)^2\]\[\sqrt{x^2+1+k}+\sqrt{x^2+2x-k}>2x+1\]

Answer 20

So, the inequality is true when k < 2x

Answer 21

Wow. Nice job deducing the factorizations.

Answer 22

Nicely done Davidc. Notice blockcolder that you can quickly deduce Davidc's answer if you write out what I first suggested and Savvy has done.