Kvant Math Problem 727
Since the perimeter is $2$, we have
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Problem
Prove the inequality $$a^2+b^2+c^2+2abc \lt 2,$$, where $a$, $b$, $c$ are the side lengths of a triangle with perimeter 2.
I. Zharov
Exploration
Since the perimeter is $2$, we have
$$a+b+c=2.$$
The expression to be bounded is
$$a^2+b^2+c^2+2abc.$$
A natural first step is to eliminate one variable using the perimeter condition. Since
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca),$$
the expression becomes
$$4-2(ab+bc+ca)+2abc.$$
The desired inequality is equivalent to
$$4-2(ab+bc+ca)+2abc<2,$$
or
$$ab+bc+ca-abc>1.$$
This reformulation looks simpler.
To test plausibility, consider the equilateral triangle $a=b=c=\frac23$. Then
$$ab+bc+ca-abc =3\cdot\frac49-\frac{8}{27} =\frac43-\frac{8}{27} =\frac{28}{27}>1.$$
For a nearly degenerate triangle, take $a=1$, $b=0.99$, $c=0.01$. Then
$$ab+bc+ca-abc =0.99+0.0099+0.01-0.0099 =1.$$
This suggests that equality is approached when the triangle becomes degenerate, which is consistent with the strict inequality required.
The expression
ab+bc+ca-abc ] resembles the identity obtained after writing $a=1-x$, $b=1-y$, $c=1-z$. Since $a+b+c=2$, we get
x+y+z=1.
$$Substituting may reveal a factorization. Computing,$$
ab+bc+ca-abc
=(1-x)(1-y)+(1-y)(1-z)+(1-z)(1-x)
-(1-x)(1-y)(1-z).
$$After expansion, everything cancels except$$
1+xyz.
This is the crucial point. Since the triangle inequalities imply $a<1$, $b<1$, $c<1$, we have $x,y,z>0$, hence $xyz>0$, giving
ab+bc+ca-abc=1+xyz>1.
That immediately yields the result. ## Problem Understanding We are given the side lengths $a,b,c$ of a triangle whose perimeter equals $2$, so
a+b+c=2.
$$We must prove$$
a^2+b^2+c^2+2abc<2.
This is a Type B problem, a pure proof. The core difficulty is finding a useful way to combine the perimeter condition with the triangle inequalities. The key observation is that every side of a triangle with perimeter $2$ is less than $1$, allowing the substitution
a=1-x,\quad b=1-y,\quad c=1-z,
with positive $x,y,z$ satisfying $x+y+z=1$. After this substitution, the target inequality reduces to the positivity of $xyz$. ## Proof Architecture First, rewrite the target inequality in terms of $ab+bc+ca-abc$ using the identity $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ and the condition $a+b+c=2$. Second, prove that each side is less than $1$; this follows from the triangle inequalities and the perimeter condition. Third, set $a=1-x$, $b=1-y$, $c=1-z$. Then $x,y,z>0$ and $x+y+z=1$. Fourth, prove the identity
ab+bc+ca-abc=1+xyz.
This follows from direct expansion and cancellation. Finally, use $xyz>0$ to obtain $ab+bc+ca-abc>1$, which is equivalent to the desired inequality. The step most likely to fail under scrutiny is the algebraic identity
ab+bc+ca-abc=1+xyz,
because an incorrect expansion would invalidate the proof. ## Solution Since
a+b+c=2,
$$we have$$
a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)
=4-2(ab+bc+ca).
$$Hence$$
a^2+b^2+c^2+2abc
=4-2(ab+bc+ca)+2abc.
$$Therefore the required inequality$$
a^2+b^2+c^2+2abc<2
$$is equivalent to$$
4-2(ab+bc+ca)+2abc<2,
$$which simplifies to$$
ab+bc+ca-abc>1.
Because $a,b,c$ are the side lengths of a triangle and
a+b+c=2,
the triangle inequality $a<b+c$ gives
a<2-a,
hence $a<1$. Similarly,
b<1,\qquad c<1.
$$Define$$
x=1-a,\qquad y=1-b,\qquad z=1-c.
$$Then$$
x>0,\qquad y>0,\qquad z>0,
$$and$$
x+y+z
=(1-a)+(1-b)+(1-c)
=3-(a+b+c)
=1.
Now [ ab+bc+ca-abc =(1-x)(1-y)+(1-y)(1-z)+(1-z)(1-x) -(1-x)(1-y)(1-z). ] Expanding the first three products, [ ab+bc+ca =3-2(x+y+z)+(xy+yz+zx). ] Since $x+y+z=1$, [ ab+bc+ca =1+(xy+yz+zx). ] Also, [ abc =(1-x)(1-y)(1-z) =1-(x+y+z)+(xy+yz+zx)-xyz. ] Using $x+y+z=1$, [ abc=(xy+yz+zx)-xyz. ] Subtracting, [ ab+bc+ca-abc = \bigl(1+(xy+yz+zx)\bigr) -\bigl((xy+yz+zx)-xyz\bigr) =1+xyz. ] Since $x,y,z>0$, [ xyz>0. ] Thus [ ab+bc+ca-abc=1+xyz>1. ] As established at the beginning, this is equivalent to [ a^2+b^2+c^2+2abc<2. ] This completes the proof. ∎ ## Verification of Key Steps The first delicate step is proving $a,b,c<1$. From $a<b+c$ and $a+b+c=2$, [ a<2-a. ] Adding $a$ to both sides gives [ 2a<2, ] hence $a<1$. The same argument applies to $b$ and $c$. If one merely used $a+b+c=2$, the conclusion $a<1$ would not follow; for example, three positive numbers summing to $2$ may include a number larger than $1$. The second delicate step is the identity [ ab+bc+ca-abc=1+xyz. ] Starting from [ ab+bc+ca =3-2(x+y+z)+(xy+yz+zx), ] and using $x+y+z=1$, we obtain [ ab+bc+ca=1+(xy+yz+zx). ] Also, [ abc =1-(x+y+z)+(xy+yz+zx)-xyz =(xy+yz+zx)-xyz. ] Subtracting yields [ 1+(xy+yz+zx)-((xy+yz+zx)-xyz) =1+xyz. ] Every term involving $xy+yz+zx$ cancels exactly once. A numerical check supports the formula. For the equilateral case $a=b=c=\frac23$, we have $x=y=z=\frac13$. Then [ 1+xyz=1+\frac1{27}=\frac{28}{27}, ] and [ ab+bc+ca-abc =\frac43-\frac8{27} =\frac{28}{27}, ] which agrees. ## Alternative Approaches Another approach uses the substitution [ a=y+z,\qquad b=z+x,\qquad c=x+y, ] where $x,y,z>0$. Since $a+b+c=2$, one gets [ x+y+z=1. ] After substituting into [ a^2+b^2+c^2+2abc, ] a straightforward expansion leads to [ 2-2xyz. ] Because $x,y,z>0$, this is strictly less than $2$, proving the claim. The main approach is shorter because it transforms the target inequality directly into [ ab+bc+ca-abc>1, ] and then reduces that expression to the simple form $1+xyz$. The amount of algebra is smaller and the role of the triangle inequalities becomes transparent.