IMO 1960 Problem 2
The expression contains the denominator
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 7m41s
Problem
For which real numbers $x$ does the following inequality hold: $$\frac{4x^2}{\left(1 - \sqrt{1+2x}\right)^2} < 2x + 9\,?$$
Exploration
The expression contains the denominator
$$\left(1-\sqrt{1+2x}\right)^2,$$
so the first task is to determine the domain. The square root requires
$$1+2x\ge 0,$$
hence
$$x\ge -\frac12.$$
However, the denominator also vanishes when
$$1-\sqrt{1+2x}=0.$$
This gives
$$\sqrt{1+2x}=1,$$
hence
$$x=0.$$
Thus the inequality is meaningful only for
$$x\ge -\frac12,\qquad x\ne 0.$$
The left-hand side has the appearance of a rationalization problem. Compute
$$1-\sqrt{1+2x} = \frac{1-(1+2x)}{1+\sqrt{1+2x}} = \frac{-2x}{1+\sqrt{1+2x}}.$$
Squaring,
$$\left(1-\sqrt{1+2x}\right)^2 = \frac{4x^2}{\left(1+\sqrt{1+2x}\right)^2}.$$
Substituting into the fraction,
$$\frac{4x^2}{(1-\sqrt{1+2x})^2} = \left(1+\sqrt{1+2x}\right)^2.$$
This simplification removes the singular-looking structure entirely.
The inequality then becomes
$$\left(1+\sqrt{1+2x}\right)^2<2x+9.$$
Expanding gives
$$1+2\sqrt{1+2x}+1+2x<2x+9,$$
hence
$$2\sqrt{1+2x}<7.$$
Since the square root is nonnegative, squaring is legitimate:
$$1+2x<\frac{49}{4},$$
so
$$x<\frac{45}{8}.$$
The tentative answer is therefore
$$-\frac12\le x<\frac{45}{8},\qquad x\ne 0.$$
The dangerous step is the rationalization identity. It divides by
$$1+\sqrt{1+2x},$$
and this quantity must never vanish. Since
$$x\ge -\frac12,$$
we have
$$\sqrt{1+2x}\ge 0,$$
thus
$$1+\sqrt{1+2x}\ge 1>0.$$
The manipulation is valid throughout the domain.
A second delicate point is squaring inequalities. Here both sides are nonnegative, because
$$2\sqrt{1+2x}\ge 0,\qquad 7>0.$$
No extraneous solutions are introduced.
Problem Understanding
The problem asks for all real numbers satisfying a given inequality involving a square root and a denominator containing the same square root. This is a classification problem, so it is Type A.
The objects involved are real numbers subject to domain restrictions coming from the square root and from division by zero. The task is to determine exactly which real numbers satisfy the inequality.
A naive approach would expand everything immediately and try to clear denominators. That leads to complicated algebra and creates a risk of introducing invalid steps when multiplying by expressions whose sign is not yet known. The key difficulty is the denominator
$$\left(1-\sqrt{1+2x}\right)^2.$$
The decisive insight is that the denominator is designed for rationalization. After rationalizing, the complicated fraction collapses into a simple quadratic expression in
$$\sqrt{1+2x}.$$
The expected answer is
$$\left[-\frac12,\frac{45}{8}\right)\setminus{0}.$$
This interval arises because the domain forces
$$x\ge -\frac12,$$
the inequality itself gives the upper bound
$$x<\frac{45}{8},$$
and the original expression is undefined at
$$x=0.$$
Proof Architecture
The proof will proceed through the following claims.
Lemma 1 states the exact domain of the original inequality:
$$x\ge -\frac12,\qquad x\ne 0.$$
This follows from the conditions for the square root to exist and for the denominator to be nonzero.
Lemma 2 states that for every number in the domain,
$$\frac{4x^2}{(1-\sqrt{1+2x})^2} = (1+\sqrt{1+2x})^2.$$
This is proved by rationalizing the denominator expression
$$1-\sqrt{1+2x}.$$
Lemma 3 states that, on the domain, the original inequality is equivalent to
$$x<\frac{45}{8}.$$
This follows by substituting the identity from Lemma 2 and solving the resulting inequality carefully.
The final step combines the domain restriction from Lemma 1 with the inequality characterization from Lemma 3. The hardest direction is preserving equivalence during algebraic manipulation, especially when squaring inequalities and simplifying denominators. The lemma most likely to fail under careless handling is Lemma 2, because an invalid rationalization could silently divide by zero.
Solution
Lemma 1
The original inequality is defined exactly for the real numbers satisfying
$$x\ge -\frac12,\qquad x\ne 0.$$
Proof
The expression
$$\sqrt{1+2x}$$
is defined if and only if
$$1+2x\ge 0.$$
Solving,
$$x\ge -\frac12.$$
The denominator
$$\left(1-\sqrt{1+2x}\right)^2$$
must be nonzero. Since a square vanishes only when its base vanishes, this requires
$$1-\sqrt{1+2x}\ne 0.$$
Hence
$$\sqrt{1+2x}\ne 1.$$
Squaring both sides gives
$$1+2x\ne 1,$$
thus
$$x\ne 0.$$
Therefore the inequality is defined exactly for
$$x\ge -\frac12,\qquad x\ne 0.$$
∎
This lemma establishes the precise domain, and omitting it would risk accepting the invalid point $x=0$.
Lemma 2
For every real number satisfying
$$x\ge -\frac12,\qquad x\ne 0,$$
one has
$$\frac{4x^2}{(1-\sqrt{1+2x})^2} = (1+\sqrt{1+2x})^2.$$
Proof
For every
$$x\ge -\frac12,$$
the quantity
$$1+\sqrt{1+2x}$$
is strictly positive, since
$$\sqrt{1+2x}\ge 0.$$
Hence division by
$$1+\sqrt{1+2x}$$
is legitimate.
Compute:
$$1-\sqrt{1+2x} = \frac{(1-\sqrt{1+2x})(1+\sqrt{1+2x})}{1+\sqrt{1+2x}}.$$
Using the identity
$$(a-b)(a+b)=a^2-b^2,$$
the numerator becomes
$$1-(1+2x)=-2x.$$
Thus
$$1-\sqrt{1+2x} = \frac{-2x}{1+\sqrt{1+2x}}.$$
Squaring both sides yields
$$(1-\sqrt{1+2x})^2 = \frac{4x^2}{(1+\sqrt{1+2x})^2}.$$
Since
$$x\ne 0,$$
the numerator
$$4x^2$$
is nonzero. Therefore division by
$$4x^2$$
is valid, and
$$\frac{4x^2}{(1-\sqrt{1+2x})^2} = (1+\sqrt{1+2x})^2.$$
∎
This lemma converts the complicated fraction into a simple square, and skipping the justification about nonzero denominators would make the argument invalid at $x=0$.
Lemma 3
On the domain
$$x\ge -\frac12,\qquad x\ne 0,$$
the original inequality is equivalent to
$$x<\frac{45}{8}.$$
Proof
By Lemma 2, the inequality
$$\frac{4x^2}{(1-\sqrt{1+2x})^2}<2x+9$$
is equivalent to
$$(1+\sqrt{1+2x})^2<2x+9.$$
Expand the left-hand side:
$$1+2\sqrt{1+2x}+(1+2x)<2x+9.$$
Combining terms gives
$$2+2x+2\sqrt{1+2x}<2x+9.$$
Subtracting
$$2x+2$$
from both sides,
$$2\sqrt{1+2x}<7.$$
Both sides are nonnegative, since
$$\sqrt{1+2x}\ge 0.$$
Therefore squaring preserves equivalence:
$$4(1+2x)<49.$$
Hence
$$4+8x<49,$$
so
$$8x<45.$$
Dividing by the positive number $8$,
$$x<\frac{45}{8}.$$
∎
This lemma determines the inequality condition itself, and a careless squaring step without checking signs could introduce false solutions.
We now determine all solutions of the problem.
By Lemma 1, the inequality is defined exactly for
$$x\ge -\frac12,\qquad x\ne 0.$$
By Lemma 3, among these numbers the inequality holds exactly when
$$x<\frac{45}{8}.$$
Combining the two conditions gives
$$-\frac12\le x<\frac{45}{8},\qquad x\ne 0.$$
It remains to verify boundary behavior explicitly.
For
$$x=-\frac12,$$
the original expression equals
$$\frac{4\left(\frac14\right)}{(1-0)^2}=1,$$
while
$$2x+9=8.$$
Hence
$$1<8,$$
so
$$x=-\frac12$$
is a solution.
For
$$x=\frac{45}{8},$$
we obtain
$$2\sqrt{1+2x}=7,$$
which gives equality rather than a strict inequality. Hence
$$x=\frac{45}{8}$$
is not a solution.
For
$$x=0,$$
the denominator vanishes, so the expression is undefined.
Therefore the complete solution set is
$$\boxed{\left[-\frac12,\frac{45}{8}\right)\setminus{0}}.$$
Verification of Key Steps
The first delicate step is the rationalization identity. Re-derive it independently:
$$(1-\sqrt{1+2x})(1+\sqrt{1+2x}) = 1-(1+2x) = -2x.$$
Hence
$$1-\sqrt{1+2x} = \frac{-2x}{1+\sqrt{1+2x}}.$$
A careless argument could fail at points where
$$1+\sqrt{1+2x}=0.$$
However,
$$\sqrt{1+2x}\ge 0,$$
so
$$1+\sqrt{1+2x}\ge 1.$$
No division by zero occurs.
The second delicate step is replacing
$$(1+\sqrt{1+2x})^2<2x+9$$
with
$$2\sqrt{1+2x}<7.$$
Expanding carefully,
$$1+2\sqrt{1+2x}+1+2x<2x+9.$$
Subtracting $2x$ from both sides yields
$$2+2\sqrt{1+2x}<9.$$
Subtracting $2$ gives
$$2\sqrt{1+2x}<7.$$
An algebraic slip in combining constants would change the final bound.
The third delicate step is squaring the inequality
$$2\sqrt{1+2x}<7.$$
Since both sides are nonnegative, squaring preserves equivalence:
$$4(1+2x)<49.$$
If one squares inequalities without checking sign conditions, false implications can arise. For example,
$$-3<2$$
but
$$9>4.$$
The nonnegativity verification is indispensable.
Alternative Approaches
A different strategy is to substitute
$$t=\sqrt{1+2x}.$$
Since
$$t\ge 0,$$
we have
$$x=\frac{t^2-1}{2}.$$
The inequality becomes
$$\frac{(t^2-1)^2}{(1-t)^2}<t^2+8.$$
Using
$$t^2-1=(t-1)(t+1),$$
the left-hand side simplifies to
$$(t+1)^2,$$
provided
$$t\ne 1.$$
The inequality then becomes
$$t^2+2t+1<t^2+8,$$
hence
$$2t<7,$$
so
$$0\le t<\frac72.$$
Transforming back,
$$0\le \sqrt{1+2x}<\frac72,$$
which yields
$$-\frac12\le x<\frac{45}{8}.$$
The condition
$$t\ne 1$$
corresponds to
$$x\ne 0.$$
This approach is shorter algebraically, but it introduces an additional variable and requires translating all domain conditions twice. The main solution keeps the original variable throughout and exposes the rationalization structure directly.