Kvant Math Problem 986
The inequality is
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m42s
Source on kvant.digital
Problem
Prove that for any positive numbers $a$ and $b$, the inequality $$2\sqrt{a\vphantom b}+3\sqrt[\scriptstyle3~]{b}\ge5\sqrt[\scriptstyle5~]{ab}.$$ holds.
Exploration
The inequality is
$$2\sqrt a+3\sqrt[3]{b}\ge 5\sqrt[5]{ab}.$$
The coefficients $2$ and $3$ add up to $5$, while the exponents are $\frac12$, $\frac13$, and $\frac15$. This suggests a weighted AM-GM inequality.
To test this idea, take $a=b=1$. Then the left-hand side equals $2+3=5$, and the right-hand side equals $5$, so equality holds.
The weighted AM-GM inequality states that for positive numbers $x,y$ and weights $\alpha,\beta>0$ with $\alpha+\beta=1$,
$$\alpha x+\beta y\ge x^\alpha y^\beta .$$
Since the coefficients are $2$ and $3$, the natural weights are
$$\alpha=\frac25,\qquad \beta=\frac35.$$
If we set
$$x=\sqrt a=a^{1/2},\qquad y=\sqrt[3]{b}=b^{1/3},$$
then
$$x^{2/5}y^{3/5} = a^{(1/2)(2/5)}b^{(1/3)(3/5)} = a^{1/5}b^{1/5} = (ab)^{1/5}.$$
This matches the right-hand side exactly. The only potentially delicate point is applying weighted AM-GM with the correct weights and checking that the resulting exponents indeed produce $(ab)^{1/5}$.
Problem Understanding
We must prove that for all positive numbers $a$ and $b$,
$$2\sqrt a+3\sqrt[3]{b}\ge 5\sqrt[5]{ab}.$$
This is a Type B problem, a pure proof.
The core difficulty is recognizing the appropriate form of the weighted arithmetic mean-geometric mean inequality and choosing variables so that the exponents combine to produce the fifth root of $ab$.
Proof Architecture
The proof uses a single lemma: for positive numbers $x,y$ and positive weights $\alpha,\beta$ satisfying $\alpha+\beta=1$,
$$\alpha x+\beta y\ge x^\alpha y^\beta.$$
This is the weighted AM-GM inequality.
We then substitute
$$x=\sqrt a,\qquad y=\sqrt[3]{b},\qquad \alpha=\frac25,\qquad \beta=\frac35,$$
and compute the resulting geometric mean.
The step most likely to fail under scrutiny is the exponent calculation
$$(a^{1/2})^{2/5}(b^{1/3})^{3/5} =(ab)^{1/5},$$
because an incorrect exponent would destroy the correspondence with the desired right-hand side.
Solution
Apply the weighted arithmetic mean-geometric mean inequality to the positive numbers
$$x=\sqrt a,\qquad y=\sqrt[3]{b}$$
with weights
$$\frac25,\qquad \frac35.$$
Since these weights are positive and sum to $1$, we have
$$\frac25,x+\frac35,y \ge x^{2/5}y^{3/5}.$$
Substituting the chosen values of $x$ and $y$ gives
$$\frac25\sqrt a+\frac35\sqrt[3]{b} \ge (\sqrt a)^{2/5}\bigl(\sqrt[3]{b}\bigr)^{3/5}.$$
Multiplying by $5$ yields
$$2\sqrt a+3\sqrt[3]{b} \ge 5(\sqrt a)^{2/5}\bigl(\sqrt[3]{b}\bigr)^{3/5}.$$
Now
$$(\sqrt a)^{2/5}\bigl(\sqrt[3]{b}\bigr)^{3/5} = a^{(1/2)(2/5)}b^{(1/3)(3/5)} = a^{1/5}b^{1/5} = (ab)^{1/5}.$$
Hence
$$2\sqrt a+3\sqrt[3]{b} \ge 5(ab)^{1/5} = 5\sqrt[5]{ab}.$$
This is exactly the required inequality.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the choice of weights. The coefficients on the left-hand side are $2$ and $3$, whose sum is $5$. Dividing by $5$ produces the weights
$$\frac25,\qquad \frac35,$$
which satisfy the condition required for weighted AM-GM.
The second delicate step is the exponent computation. Starting from
$$(\sqrt a)^{2/5}\bigl(\sqrt[3]{b}\bigr)^{3/5},$$
we obtain
$$(a^{1/2})^{2/5}=a^{1/5}, \qquad (b^{1/3})^{3/5}=b^{1/5}.$$
Multiplying gives
$$a^{1/5}b^{1/5}=(ab)^{1/5}.$$
No additional assumptions are used beyond the positivity of $a$ and $b$.
A numerical check confirms the structure. For $a=32$ and $b=243$,
$$\sqrt a\approx5.657,\qquad \sqrt[3]{b}\approx6.240,$$
so
$$2\sqrt a+3\sqrt[3]{b}\approx30.034.$$
Meanwhile,
$$5\sqrt[5]{ab} = 5\sqrt[5]{7776} \approx29.985.$$
The inequality holds, consistent with the proof.
Alternative Approaches
The inequality can also be obtained from the general weighted AM-GM statement written directly in the form
$$mX+nY\ge (m+n)X^{m/(m+n)}Y^{n/(m+n)}$$
for positive $X,Y$ and positive numbers $m,n$. Taking
$$m=2,\qquad n=3,\qquad X=\sqrt a,\qquad Y=\sqrt[3]{b},$$
gives
$$2\sqrt a+3\sqrt[3]{b} \ge 5(\sqrt a)^{2/5}\bigl(\sqrt[3]{b}\bigr)^{3/5} = 5\sqrt[5]{ab}.$$
This is essentially the same idea expressed in a more specialized form. The weighted AM-GM approach is preferable because it reveals immediately why the coefficients $2$ and $3$ and the exponents $\frac12$, $\frac13$, and $\frac15$ fit together exactly.