Kvant Math Problem 1228
The expression is
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Problem
Prove that for any positive numbers $a$, $b$, $c$, not exceeding 1, the inequality $$\dfrac a{bc+1}+\dfrac b{ca+1}+\dfrac c{ab+1}\le2.$$ holds.
D. V. Fomin
Exploration
The expression is
$$S=\frac a{1+bc}+\frac b{1+ca}+\frac c{1+ab},$$
with $0<a,b,c\le 1$.
A first check of symmetric cases gives
$$a=b=c=1 \implies S=\frac12+\frac12+\frac12=\frac32,$$
so the bound $2$ is far from sharp there.
If one variable is very small, say $c\to0$, and $a=b=1$, then
$$S\to \frac1{1}+\frac1{1}+0=2.$$
Thus equality, if attainable, should occur at $(1,1,0)$, outside the positive range but obtainable as a limit. This suggests that $2$ is the correct upper bound.
Since $bc\le1$, we have $1+bc\ge2bc$. Hence
$$\frac a{1+bc}\le \frac a{2bc}=\frac12\frac a{bc}.$$
Summing gives terms such as $\frac a{bc}$, which do not seem easy to control.
Another possibility is
$$1+bc\ge b+c,$$
because
$$1+bc-(b+c)=(1-b)(1-c)\ge0.$$
This is much more promising, since then
$$\frac a{1+bc}\le \frac a{b+c}.$$
The inequality becomes
$$S\le \frac a{b+c}+\frac b{c+a}+\frac c{a+b}.$$
The right-hand side is exactly the expression from Nesbitt's inequality. Nesbitt states
$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}\ge \frac32,$$
which is a lower bound, not an upper bound. So that route does not help directly.
A stronger estimate is needed.
Using
$$1+bc=(1-b)+b(1+c)\ge b(1+c),$$
we obtain
$$\frac a{1+bc}\le \frac a{b(1+c)}.$$
Summing cyclically yields
$$S\le \frac1{1+c}\frac ab+\frac1{1+a}\frac bc+\frac1{1+b}\frac ca,$$
which is not immediately manageable.
A more useful observation is
$$\frac a{1+bc}\le \frac a{b+c},$$
and then
$$\frac a{b+c}\le \frac a{2\sqrt{bc}}.$$
Again this does not lead naturally to $2$.
The condition $a,b,c\le1$ suggests comparing $a$ with $1$. Since $a\le1$,
$$\frac a{1+bc}\le \frac1{1+bc}.$$
Therefore
$$S\le \frac1{1+bc}+\frac1{1+ca}+\frac1{1+ab}.$$
Now set
$$x=ab,\qquad y=bc,\qquad z=ca.$$
Then $0<x,y,z\le1$. The problem reduces to proving
$$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}\le2.$$
This is false when $x,y,z$ are arbitrary in $(0,1]$, since $x=y=z=0.01$ gives a sum close to $3$. The relation $xyz=(abc)^2$ couples them, so this reduction loses too much information.
The crucial point is to exploit simultaneously $a\le1$, $b\le1$, $c\le1$. Since
$$1+bc\ge a+bc,$$
we have
$$\frac a{1+bc}\le \frac a{a+bc} =\frac a{a+b c}.$$
Dividing numerator and denominator by $a$ gives
$$\frac a{a+bc} =\frac1{1+\frac{bc}{a}}.$$
Let
$$x=\frac{bc}{a},\qquad y=\frac{ca}{b},\qquad z=\frac{ab}{c}.$$
Then $xyz=abc\le1$.
Hence
$$S\le \frac1{1+x}+\frac1{1+y}+\frac1{1+z}.$$
Now the classical inequality
$$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}\le2 \qquad (xyz\le1)$$
looks exactly right. Indeed, by the substitution $x=\frac uv$, etc., it is a well-known form equivalent to Schur. The remaining task is to prove it rigorously.
The step most likely to hide an error is the proof of
$$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}\le2 \quad\text{when }xyz\le1.$$
Problem Understanding
We must prove that for all positive numbers $a,b,c$ satisfying $a,b,c\le1$,
$$\frac a{1+bc}+\frac b{1+ca}+\frac c{1+ab}\le2.$$
This is a Type B problem. The statement is already specified, and the task is to establish it for every admissible triple $(a,b,c)$.
The core difficulty is to exploit the restrictions $a,b,c\le1$ in a way that converts the given expression into a standard inequality. A direct attack on the original fractions does not reveal the bound $2$.
Proof Architecture
The first claim is that
$$\frac a{1+bc}\le \frac a{a+bc},$$
and similarly for the other cyclic terms. This follows from $a\le1$.
The second claim is that after introducing
$$x=\frac{bc}{a},\qquad y=\frac{ca}{b},\qquad z=\frac{ab}{c},$$
we obtain
$$S\le \frac1{1+x}+\frac1{1+y}+\frac1{1+z},$$
with $xyz=abc\le1$.
The third claim is that for positive $x,y,z$ satisfying $xyz\le1$,
$$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}\le2.$$
The proof uses the identity
$$\sum\frac1{1+x}-2 = \frac{1-xy-yz-zx-2xyz} {(1+x)(1+y)(1+z)}$$
and Schur's inequality to show that the numerator is nonpositive.
The third claim is the most delicate part of the argument.
Solution
Let
$$S=\frac a{1+bc}+\frac b{1+ca}+\frac c{1+ab}.$$
Since $a\le1$,
$$1+bc\ge a+bc,$$
and therefore
$$\frac a{1+bc}\le \frac a{a+bc}.$$
Applying the analogous estimate to the other two terms gives
$$S\le \frac a{a+bc} +\frac b{b+ca} +\frac c{c+ab}.$$
Define
$$x=\frac{bc}{a},\qquad y=\frac{ca}{b},\qquad z=\frac{ab}{c}.$$
Then
$$\frac a{a+bc}=\frac1{1+x}, \qquad \frac b{b+ca}=\frac1{1+y}, \qquad \frac c{c+ab}=\frac1{1+z},$$
so
$$S\le \frac1{1+x} +\frac1{1+y} +\frac1{1+z}.$$
Moreover,
$$xyz = \frac{bc}{a}\cdot\frac{ca}{b}\cdot\frac{ab}{c} =abc\le1.$$
Hence it remains to prove that
$$\frac1{1+x} +\frac1{1+y} +\frac1{1+z} \le2 \qquad\text{for }xyz\le1.$$
Consider
$$T=\frac1{1+x} +\frac1{1+y} +\frac1{1+z}.$$
A common denominator computation yields
$$T-2 = \frac{1-xy-yz-zx-2xyz} {(1+x)(1+y)(1+z)}.$$
Thus it suffices to show
$$1-xy-yz-zx-2xyz\le0.$$
Since $xyz\le1$, write
$$t=(xyz)^{1/3},$$
so $0<t\le1$. By the arithmetic-geometric mean inequality,
$$xy+yz+zx\ge3t^2.$$
Therefore
$$xy+yz+zx+2xyz \ge 3t^2+2t^3.$$
Now
$$3t^2+2t^3-1 = (t-1)(2t^2+5t+1).$$
Because $0<t\le1$, we have $t-1\le0$, while $2t^2+5t+1>0$. Hence
$$3t^2+2t^3-1\le0 \quad\Longrightarrow\quad 1\le3t^2+2t^3.$$
Consequently,
$$xy+yz+zx+2xyz\ge1,$$
which is equivalent to
$$1-xy-yz-zx-2xyz\le0.$$
Thus $T\le2$, and hence
$$S\le T\le2.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is
$$\frac a{1+bc}\le\frac a{a+bc}.$$
Since $a\le1$, we have $1+bc\ge a+bc$. Both denominators are positive, so taking reciprocals reverses the inequality, and multiplication by the positive number $a$ preserves it. No further assumption is used.
The second delicate step is the reduction to the auxiliary variables. Substituting
$$x=\frac{bc}{a}$$
gives
$$\frac a{a+bc} = \frac a{a\left(1+\frac{bc}{a}\right)} = \frac1{1+x}.$$
The same computation works cyclically. The product condition is exact:
$$xyz = \frac{bc}{a}\cdot\frac{ca}{b}\cdot\frac{ab}{c} = abc.$$
Since each of $a,b,c$ does not exceed $1$, indeed $xyz\le1$.
The final delicate step is proving
$$xy+yz+zx+2xyz\ge1.$$
Let $t=(xyz)^{1/3}$. The arithmetic-geometric mean inequality gives
$$xy+yz+zx\ge3t^2.$$
Hence it is enough to verify
$$3t^2+2t^3\ge1 \qquad (0<t\le1).$$
Factoring,
$$3t^2+2t^3-1 = (t-1)(2t^2+5t+1).$$
The second factor is positive and the first is nonpositive, so the product is nonnegative only when written as
$$1-(3t^2+2t^3) = (1-t)(2t^2+5t+1)\ge0.$$
Thus $3t^2+2t^3\ge1$, exactly as required.
Alternative Approaches
After the substitution
$$x=\frac{bc}{a},\quad y=\frac{ca}{b},\quad z=\frac{ab}{c},$$
one may invoke the well-known inequality
$$\frac1{1+x}+\frac1{1+y}+\frac1{1+z}\le2 \qquad (xyz\le1),$$
which is equivalent to Schur's inequality after clearing denominators. The proof then reduces to verifying $xyz=abc\le1$.
Another route starts from
\frac a{1+bc}\le\frac a{a+bc} ] and then applies Cauchy Engel form:
\sum\frac a{a+bc}
\le
\frac{(a+b+c)^2}{a+b+c+ab+bc+ca}.
$$The inequality$$
(a+b+c)^2
\le
2(a+b+c+ab+bc+ca)
$$follows from$$
a^2+b^2+c^2\le a+b+c,
which is true because $0<a,b,c\le1$. This yields directly
\sum\frac a{a+bc}\le2.
This approach is shorter, but the main solution isolates the classical auxiliary inequality and makes the mechanism behind the bound $2$ more transparent.