Kvant Math Problem 647
The inequality is symmetric in $a$ and $b$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m42s
Source on kvant.digital
Problem
Prove that for all $a\ge\dfrac12$, $b\ge\dfrac12$ the inequality $$\left(\dfrac{a^2-b^2}{2}\right)^2\ge\sqrt{\dfrac{a^2+b^2}{2}}-\dfrac{a+b}{2}.$$ holds.
S. V. Fomin
Exploration
The inequality is symmetric in $a$ and $b$. The right-hand side contains the classical comparison between the root mean square and the arithmetic mean:
$$\sqrt{\frac{a^2+b^2}{2}}\ge \frac{a+b}{2}.$$
Hence the right-hand side is nonnegative.
The left-hand side equals
$$\frac{(a^2-b^2)^2}{4} =\frac{(a-b)^2(a+b)^2}{4}.$$
Since $a,b\ge \tfrac12$, we have $a+b\ge1$, so
$$\frac{(a^2-b^2)^2}{4}\ge \frac{(a-b)^2}{4}.$$
This suggests comparing the quantity
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}$$
with $\frac{(a-b)^2}{4}$.
Trying small examples:
For $a=b$, both sides are $0$.
For $a=1$, $b=\tfrac12$,
$$\sqrt{\frac{1+\frac14}{2}}-\frac34 =\sqrt{\frac58}-\frac34 \approx0.0406,$$
while
$$\frac{(a-b)^2}{4}=\frac1{16}=0.0625.$$
The comparison works.
The expression
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}$$
can be rationalized:
$$\frac{\frac{a^2+b^2}{2}-\frac{(a+b)^2}{4}} {\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}} = \frac{(a-b)^2} {4\left(\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}\right)}.$$
Thus it suffices to show
$$\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}\ge1.$$
Since $a+b\ge1$,
$$\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}\ge \frac{a+b}{2}+\frac{a+b}{2}=a+b\ge1.$$
This gives exactly the desired estimate. The potentially delicate point is keeping track of the direction of inequalities after replacing the left-hand side by the weaker bound $\frac{(a-b)^2}{4}$.
Problem Understanding
We must prove that for all $a,b\ge \tfrac12$,
$$\left(\frac{a^2-b^2}{2}\right)^2 \ge \sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}.$$
This is a Type B problem, a pure proof.
The core difficulty is to relate the difference between the root mean square and the arithmetic mean on the right-hand side to the factor $(a-b)^2$ appearing naturally in the left-hand side.
The condition $a,b\ge \tfrac12$ suggests using $a+b\ge1$, which is exactly the extra information needed after rationalizing the right-hand side.
Proof Architecture
First, prove that
$$\left(\frac{a^2-b^2}{2}\right)^2\ge \frac{(a-b)^2}{4},$$
because $a+b\ge1$.
Next, rewrite
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}$$
by rationalization as
$$\frac{(a-b)^2} {4\left(\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}\right)}.$$
Then show that
$$\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}\ge1.$$
This follows from the RMS-AM inequality together with $a+b\ge1$.
Finally compare the obtained upper bound for the right-hand side with $\frac{(a-b)^2}{4}$ and conclude.
The most delicate lemma is the rationalized form of the right-hand side, because an algebraic error there would invalidate the comparison.
Solution
Since $a,b\ge \tfrac12$, we have
$$a+b\ge1.$$
Therefore
$$\left(\frac{a^2-b^2}{2}\right)^2 = \frac{(a-b)^2(a+b)^2}{4} \ge \frac{(a-b)^2}{4}. \tag{1}$$
Now consider the right-hand side. Rationalizing gives
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2} = \frac{\frac{a^2+b^2}{2}-\frac{(a+b)^2}{4}} {\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}}.$$
The numerator equals
$$\frac{2(a^2+b^2)-(a+b)^2}{4} = \frac{a^2-2ab+b^2}{4} = \frac{(a-b)^2}{4}.$$
Hence
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2} = \frac{(a-b)^2} {4\left(\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}\right)}. \tag{2}$$
By the RMS-AM inequality,
$$\sqrt{\frac{a^2+b^2}{2}} \ge \frac{a+b}{2}.$$
Consequently,
$$\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2} \ge a+b \ge 1.$$
Using this in (2), we obtain
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2} \le \frac{(a-b)^2}{4}. \tag{3}$$
Combining (1) and (3) yields
$$\left(\frac{a^2-b^2}{2}\right)^2 \ge \frac{(a-b)^2}{4} \ge \sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2}.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is
$$\frac{(a-b)^2(a+b)^2}{4}\ge \frac{(a-b)^2}{4}.$$
The factor $(a-b)^2$ is nonnegative. Since $a,b\ge \tfrac12$, we have $(a+b)^2\ge1$. Multiplying a nonnegative quantity by a factor at least $1$ preserves the inequality.
The second delicate step is the rationalization. Computing the numerator directly,
$$\frac{a^2+b^2}{2}-\frac{(a+b)^2}{4} = \frac{2a^2+2b^2-a^2-2ab-b^2}{4} = \frac{(a-b)^2}{4}.$$
Any sign mistake here would produce $\frac{(a+b)^2}{4}$ instead and destroy the argument.
The third delicate step is proving that the denominator in (2) is at least $1$. RMS-AM gives
$$\sqrt{\frac{a^2+b^2}{2}}\ge \frac{a+b}{2},$$
hence the denominator is at least $a+b$. The hypothesis $a,b\ge \tfrac12$ is used exactly here to obtain $a+b\ge1$. Without this hypothesis the statement would generally fail.
Alternative Approaches
One may start from the identity
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2} = \frac{(a-b)^2} {4\left(\sqrt{\frac{a^2+b^2}{2}}+\frac{a+b}{2}\right)}$$
and immediately estimate the denominator from below by $a+b$. This yields
$$\sqrt{\frac{a^2+b^2}{2}}-\frac{a+b}{2} \le \frac{(a-b)^2}{4(a+b)}.$$
Since $a+b\ge1$,
$$\frac{(a-b)^2}{4(a+b)} \le \frac{(a-b)^2(a+b)^2}{4} = \left(\frac{a^2-b^2}{2}\right)^2,$$
which proves the claim.
This approach is essentially equivalent to the main proof, but it packages the comparison into a single estimate after rationalization. The main proof separates the argument into two transparent bounds, making the role of the condition $a+b\ge1$ especially clear.