Kvant Math Problem 579
For $n=1$ the inequality becomes
Verified: yes
Verdicts: PASS + PASS
Solve time: 15m32s
Source on kvant.digital
Problem
Prove that for any numbers $x_1$, $x_2$, $\ldots$, $x_n$ belonging to the interval $[0; 1]$, the inequality $$(x_1+x_2+\ldots+x_n+1)^2 \ge 4(x_1^2 + x_2^2+\ldots+x_n^2).$$ holds.
A. Plotkin, S. V. Fomin
All-Union Mathematical Olympiad for School Students (1979, Grade 9)
Exploration
For $n=1$ the inequality becomes
$$(x_1+1)^2\ge 4x_1^2.$$
Since $x_1\in[0,1]$,
$$(x_1+1)^2-4x_1^2=(1-x_1)(1+3x_1)\ge0.$$
For $n=2$,
$$(x_1+x_2+1)^2\ge 4(x_1^2+x_2^2).$$
Testing several values gives equality at $(1,0)$ and $(0,1)$:
$$(1+0+1)^2=4=4(1^2+0^2).$$
At $(1,1)$ the left side is $9$ and the right side is $8$.
The interval condition $0\le x_i\le1$ suggests comparing $x_i^2$ with $x_i$. Since $x_i^2\le x_i$, we have
$$\sum x_i^2\le \sum x_i.$$
Let
$$S=x_1+\cdots+x_n.$$
Then it would suffice to prove
$$(S+1)^2\ge 4S.$$
Indeed,
$$4\sum x_i^2\le 4S.$$
The remaining inequality is
$$(S+1)^2-4S=(S-1)^2\ge0.$$
This works immediately.
The step most likely to hide an error is replacing $\sum x_i^2$ by $S$. One must check that the inequality direction is preserved after multiplication by $4$, and that the resulting stronger statement indeed implies the original claim.
Problem Understanding
We are given numbers $x_1,\dots,x_n$ in the interval $[0,1]$ and must prove
$$(x_1+\cdots+x_n+1)^2\ge 4(x_1^2+\cdots+x_n^2).$$
This is a Type B problem, a pure proof.
The core difficulty is to exploit the restriction $0\le x_i\le1$. That restriction immediately yields $x_i^2\le x_i$, reducing the problem to a one-variable inequality in the sum $S=x_1+\cdots+x_n$.
Proof Architecture
Let $S=x_1+\cdots+x_n$.
Lemma 1. For every $i$, $x_i^2\le x_i$; this follows from $0\le x_i\le1$.
Lemma 2. Summing the inequalities of Lemma 1 gives
$$x_1^2+\cdots+x_n^2\le S.$$
Lemma 3. The inequality
$$(S+1)^2\ge 4S$$
holds for every real $S$; this is equivalent to
$$(S-1)^2\ge0.$$
Combining Lemmas 2 and 3 yields
$$(S+1)^2\ge4S\ge4(x_1^2+\cdots+x_n^2),$$
which is exactly the desired inequality.
The most delicate point is Lemma 2, because the entire argument depends on replacing the sum of squares by the sum itself with the correct inequality direction.
Solution
Set
$$S=x_1+x_2+\cdots+x_n.$$
Since each $x_i$ belongs to the interval $[0,1]$, we have
$$x_i^2\le x_i$$
for every $i$.
Adding these inequalities for $i=1,2,\dots,n$, we obtain
$$x_1^2+x_2^2+\cdots+x_n^2\le x_1+x_2+\cdots+x_n=S.$$
Multiplying by $4$ gives
$$4(x_1^2+x_2^2+\cdots+x_n^2)\le 4S.$$
It remains to compare $(S+1)^2$ with $4S$. We compute
$$(S+1)^2-4S=S^2-2S+1=(S-1)^2.$$
Since a square is nonnegative,
$$(S+1)^2\ge4S.$$
Combining the last inequality with
$$4S\ge4(x_1^2+x_2^2+\cdots+x_n^2),$$
we obtain
$$(S+1)^2\ge4(x_1^2+x_2^2+\cdots+x_n^2).$$
Substituting back $S=x_1+\cdots+x_n$ yields
$$(x_1+x_2+\cdots+x_n+1)^2 \ge 4(x_1^2+x_2^2+\cdots+x_n^2).$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the inequality $x_i^2\le x_i$. From $0\le x_i\le1$ we have
$$x_i(1-x_i)\ge0.$$
Rearranging gives
$$x_i-x_i^2\ge0,$$
hence
$$x_i^2\le x_i.$$
If the condition $x_i\le1$ were omitted, the statement would fail. For example, $x_i=2$ gives $x_i^2=4>x_i=2$.
The second delicate step is passing from
$$x_1^2+\cdots+x_n^2\le S$$
to
$$4(x_1^2+\cdots+x_n^2)\le4S.$$
The factor $4$ is positive, so multiplication preserves the inequality direction. Using a negative factor would reverse the direction and invalidate the argument.
The final step is
$$(S+1)^2-4S=(S-1)^2.$$
Expanding confirms
$$S^2+2S+1-4S=S^2-2S+1.$$
Since $(S-1)^2\ge0$ for every real $S$, the inequality holds without any additional assumptions on $S$.
Alternative Approaches
Starting from the desired inequality, move the right-hand side to the left:
$$(x_1+\cdots+x_n+1)^2-4\sum_{i=1}^n x_i^2.$$
Let $S=\sum x_i$. Then
$$(S+1)^2-4\sum x_i^2 =(S-1)^2+4\Bigl(S-\sum x_i^2\Bigr).$$
Because $x_i^2\le x_i$ for each $i$,
$$S-\sum x_i^2=\sum (x_i-x_i^2)\ge0.$$
Both terms on the right-hand side are nonnegative, so the whole expression is nonnegative.
This proof is essentially equivalent to the main one, but it exhibits the left-hand side directly as a sum of nonnegative quantities. The main approach is shorter because it reduces the problem immediately to the elementary inequality $(S-1)^2\ge0$.