Kvant Math Problem 1067
Let
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m36s
Source on kvant.digital
Problem
Prove that for the nonnegative numbers $x$, $y$, $z$ satisfying the condition $x^2+y^2+z^2=1$, the inequality $$\dfrac x{1-x^2}+\dfrac y{1-y^2}+\dfrac z{1-z^2}\ge\dfrac{3\sqrt3}2.$$ holds.
V. E. Matizen
Exploration
Let
$$f(t)=\frac{t}{1-t^2}, \qquad 0\le t<1.$$
The condition is
$$x^2+y^2+z^2=1,$$
and we must prove
$$f(x)+f(y)+f(z)\ge \frac{3\sqrt3}{2}.$$
The right-hand side suggests checking the symmetric point
$$x=y=z=\frac1{\sqrt3}.$$
Then
$$f!\left(\frac1{\sqrt3}\right) = \frac{1/\sqrt3}{1-1/3} = \frac{\sqrt3}{2},$$
hence
$$3f!\left(\frac1{\sqrt3}\right)=\frac{3\sqrt3}{2}.$$
So equality is attained there.
A natural conjecture is that Jensen's inequality applies. Writing
$$u=x^2,\quad v=y^2,\quad w=z^2,$$
we have
$$u+v+w=1$$
and
$$f(x)=\frac{\sqrt u}{1-u}.$$
Define
$$g(u)=\frac{\sqrt u}{1-u}.$$
If $g$ is convex on $[0,1)$, then
$$g(u)+g(v)+g(w)\ge 3g!\left(\frac{u+v+w}{3}\right) =3g!\left(\frac13\right),$$
which is exactly the desired inequality.
The crucial point is the convexity of $g$. Compute:
$$g'(u)=\frac{1+u}{2\sqrt u(1-u)^2},$$
and after simplification
$$g''(u) = \frac{3u^2+6u-1}{4u^{3/2}(1-u)^3}.$$
This is not always nonnegative, since at small $u$ the numerator is negative. Thus global convexity fails.
A different idea is needed.
Observe that
$$\frac{t}{1-t^2} = \frac12\left(\frac1{1-t}-\frac1{1+t}\right).$$
That does not interact simply with the condition.
Instead, compare $f(t)$ with a linear function of $t$. Since equality should occur at $t=1/\sqrt3$, perhaps
$$f(t)\ge at$$
with equality at $t=1/\sqrt3$. We need
$$a,(x+y+z)\ge \frac{3\sqrt3}{2}.$$
Because
$$x+y+z\ge \sqrt{x^2+y^2+z^2}=1,$$
this would require $a\ge 3\sqrt3/2$.
Check
$$f(t)-\frac{3\sqrt3}{2}t = t\left(\frac1{1-t^2}-\frac{3\sqrt3}{2}\right).$$
The factor in parentheses vanishes at
$$1-t^2=\frac2{3\sqrt3},$$
not at $t=1/\sqrt3$, so this is not promising.
A better comparison is with
$$\frac32,t.$$
Then
$$f(t)-\frac32 t = \frac{t(3t^2-1)}{2(1-t^2)}.$$
Summing yields
$$\sum f(x)-\frac32(x+y+z) = \frac12\sum \frac{x(3x^2-1)}{1-x^2}.$$
Not immediately useful.
The tangent-line method at the equality point is more promising. Let
$$h(u)=\frac{\sqrt u}{1-u}.$$
Since $u+v+w=1$, it would suffice to prove
$$h(u)\ge h!\left(\frac13\right) +h'!\left(\frac13\right)\left(u-\frac13\right)$$
for all $u\in[0,1)$.
That is, the tangent at $u=1/3$ must lie below the whole graph. This is weaker than convexity and may still hold.
Now
$$h!\left(\frac13\right)=\frac{\sqrt3}{2}, \qquad h'!\left(\frac13\right)=\frac{3\sqrt3}{2}.$$
Hence we need
$$\frac{\sqrt u}{1-u} \ge \frac{\sqrt3}{2} +\frac{3\sqrt3}{2}\left(u-\frac13\right) = \frac{3\sqrt3}{2}u.$$
Remarkably, the constant terms cancel. Thus it remains to prove
$$\frac{\sqrt u}{1-u}\ge \frac{3\sqrt3}{2}u.$$
Let $s=\sqrt u$. Then
$$\frac{s}{1-s^2}\ge \frac{3\sqrt3}{2}s^2.$$
For $s>0$ this becomes
$$2\ge 3\sqrt3,s(1-s^2).$$
The maximum of $s(1-s^2)$ on $[0,1]$ occurs at
$$1-3s^2=0, \qquad s=\frac1{\sqrt3},$$
and equals
$$\frac2{3\sqrt3}.$$
Thus the inequality holds exactly, with equality at $s=1/\sqrt3$.
This yields
$$f(x)+f(y)+f(z) \ge \frac{3\sqrt3}{2}(x^2+y^2+z^2) = \frac{3\sqrt3}{2}.$$
The proof is surprisingly short.
Problem Understanding
We are given nonnegative numbers $x,y,z$ satisfying
$$x^2+y^2+z^2=1,$$
and must prove
$$\frac{x}{1-x^2} +\frac{y}{1-y^2} +\frac{z}{1-z^2} \ge \frac{3\sqrt3}{2}.$$
This is a Type B problem, a pure inequality proof.
The main difficulty is to find a lower bound for each term that is compatible with the constraint involving the squares $x^2,y^2,z^2$. Since the equality case occurs at
$$x=y=z=\frac1{\sqrt3},$$
the lower bound should be tangent to the function
$$t\mapsto \frac{t}{1-t^2}$$
at that point.
Proof Architecture
Let
$$f(t)=\frac{t}{1-t^2}.$$
We first prove that for every $t\in[0,1)$,
$$f(t)\ge \frac{3\sqrt3}{2},t^2.$$
This follows by showing that the equivalent inequality
$$2\ge 3\sqrt3,t(1-t^2)$$
holds because the maximum of $t(1-t^2)$ on $[0,1]$ equals $2/(3\sqrt3)$.
Next, apply this estimate to $x,y,z$ and sum:
$$\sum f(x)\ge \frac{3\sqrt3}{2}\sum x^2.$$
Using
$$x^2+y^2+z^2=1,$$
we obtain the required inequality.
The potentially delicate point is proving the sharp bound
$$t(1-t^2)\le \frac{2}{3\sqrt3}.$$
Solution
Define
$$f(t)=\frac{t}{1-t^2}, \qquad 0\le t<1.$$
We claim that
$$f(t)\ge \frac{3\sqrt3}{2},t^2 \qquad (0\le t<1).$$
Since $1-t^2>0$, this inequality is equivalent to
$$t\ge \frac{3\sqrt3}{2}t^2(1-t^2).$$
For $t=0$ the inequality is immediate. Assume $t>0$ and divide by $t$:
$$1\ge \frac{3\sqrt3}{2}t(1-t^2).$$
Thus it suffices to prove
$$t(1-t^2)\le \frac{2}{3\sqrt3}.$$
Consider
$$\phi(t)=t(1-t^2)=t-t^3$$
on the interval $[0,1]$. Its derivative is
$$\phi'(t)=1-3t^2.$$
Hence
$$\phi'(t)=0 \iff t=\frac1{\sqrt3}.$$
Since $\phi'(t)>0$ for $0\le t<1/\sqrt3$ and $\phi'(t)<0$ for $1/\sqrt3<t\le1$, the function attains its maximum at $t=1/\sqrt3$. The maximal value is
$$\phi!\left(\frac1{\sqrt3}\right) = \frac1{\sqrt3} \left(1-\frac13\right) = \frac{2}{3\sqrt3}.$$
Therefore
$$t(1-t^2)\le \frac{2}{3\sqrt3},$$
which proves
$$\frac{t}{1-t^2}\ge \frac{3\sqrt3}{2},t^2.$$
Applying this estimate to $x,y,z$ and summing, we obtain
$$\frac{x}{1-x^2} +\frac{y}{1-y^2} +\frac{z}{1-z^2} \ge \frac{3\sqrt3}{2}(x^2+y^2+z^2).$$
Using the condition
$$x^2+y^2+z^2=1,$$
we get
$$\frac{x}{1-x^2} +\frac{y}{1-y^2} +\frac{z}{1-z^2} \ge \frac{3\sqrt3}{2}.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the passage
$$\frac{t}{1-t^2}\ge \frac{3\sqrt3}{2}t^2 \iff t(1-t^2)\le \frac{2}{3\sqrt3}.$$
Because $1-t^2>0$ for $0\le t<1$, multiplication by $1-t^2$ preserves the direction of the inequality. When $t=0$, division by $t$ is not used; that case is checked separately.
The second delicate step is finding the maximum of
$$\phi(t)=t-t^3.$$
The derivative is
$$\phi'(t)=1-3t^2.$$
There is a unique critical point in $[0,1]$, namely $t=1/\sqrt3$. Since the derivative changes from positive to negative there, this point gives the global maximum. Evaluating yields
$$\phi!\left(\frac1{\sqrt3}\right) = \frac{2}{3\sqrt3}.$$
A careless argument that merely identifies the critical point without checking the sign change could miss the fact that one must obtain the global maximum on the whole interval.
The equality case is also consistent. Equality in
$$\frac{t}{1-t^2}\ge \frac{3\sqrt3}{2}t^2$$
holds precisely when
$$t=\frac1{\sqrt3}.$$
Hence equality in the original inequality occurs when
$$x=y=z=\frac1{\sqrt3},$$
which satisfies the constraint.
Alternative Approaches
One may set
$$u=x^2,\quad v=y^2,\quad w=z^2,$$
and write
$$\frac{x}{1-x^2} = \frac{\sqrt u}{1-u}.$$
The tangent line to the function
$$h(u)=\frac{\sqrt u}{1-u}$$
at $u=1/3$ is
$$h(u)\ge \frac{3\sqrt3}{2}u.$$
Summing this inequality for $u,v,w$ immediately gives
$$\sum \frac{\sqrt u}{1-u} \ge \frac{3\sqrt3}{2}(u+v+w) = \frac{3\sqrt3}{2}.$$
The proof above is preferable because it avoids introducing new variables and reduces the problem to a single elementary maximization of the cubic function $t-t^3$.