Let m be the minimum possible value of log₃ (3^y₁ + 3^y₂ + 3^y₃), where y₁, y₂, y₃ are real numbers

Q: Wait, why can we use AM-GM here? Are the terms always positive?

Excellent question! The AM-GM inequality strictly requires positive real numbers. In the first part, $3^{y_1}, 3^{y_2}, 3^{y_3}$ are exponential functions, which are always positive for any real $y_i$. In the second part, the problem explicitly states that $x_1, x_2, x_3$ are positive real numbers. So, the conditions are perfectly satisfied!

Q: Why does the inequality sign stay the same when we take the logarithm?

That depends on the base! We are using $\log_3$, and since the base $3 > 1$, the logarithmic function is strictly increasing. This means if $A \ge B$, then $\log_3(A) \ge \log_3(B)$. If the base were between $0$ and $1$, the inequality would flip. Always check your base!

Q: Can I use Lagrange Multipliers instead of AM-GM?

You absolutely can! Lagrange Multipliers are a powerful, general method for constrained optimization. However, for symmetric expressions like these, AM-GM is often much faster and more elegant. In a high-pressure exam like JEE Advanced, speed and accuracy are key, so mastering the AM-GM shortcut is a huge advantage.

Question

Let $m$ be the minimum possible value of $\log_3 (3^{y_1} + 3^{y_2} + 3^{y_3})$, where $y_1, y_2, y_3$ are real numbers for which $y_1 + y_2 + y_3 = 9$. Let $M$ be the maximum possible value of $(\log_3 x_1 + \log_3 x_2 + \log_3 x_3)$, where $x_1, x_2, x_3$ are positive real numbers for which $x_1 + x_2 + x_3 = 9$. Then the value of $\log_2 (m^3) + \log_3 (M^2)$ is ______.

Sigma Percentile · Accepted Answer

### The Symphony of Inequalities: A Journey into Optimization

Welcome, warriors of JEE! Today, we are not just solving an algebra problem; we are embarking on a journey to understand the elegant balance of numbers. We are going to explore the AM-GM inequality, a fundamental pillar of mathematics that bridges the gap between arithmetic and geometry.

### Phase 1: The Exponential Challenge

Our first mission is to find the minimum value $m$ of the expression $\log_3 (3^{y_1} + 3^{y_2} + 3^{y_3})$, given the constraint $y_1 + y_2 + y_3 = 9$.

The presence of a logarithm with base $3$ is our first clue. Since $3 > 1$, the function $f(t) = \log_3(t)$ is strictly increasing. This means that to minimize the entire expression, we simply need to minimize the argument inside the logarithm: $S = 3^{y_1} + 3^{y_2} + 3^{y_3}$.

Here, the AM-GM inequality becomes our most trusted ally. For any positive real numbers $a, b, c$, we know that $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$. Let us apply this to our terms $3^{y_1}, 3^{y_2}, 3^{y_3}$:

$$\frac{3^{y_1} + 3^{y_2} + 3^{y_3}}{3} \ge \sqrt[3]{3^{y_1} \cdot 3^{y_2} \cdot 3^{y_3}}$$

When we multiply these exponential terms, we use the property $a^m \cdot a^n = a^{m+n}$. Thus, the product becomes $3^{y_1 + y_2 + y_3}$.

Substituting our constraint $y_1 + y_2 + y_3 = 9$, the geometric mean simplifies beautifully to $\sqrt[3]{3^9} = 3^3 = 27$. Therefore, we have:

$$\frac{S}{3} \ge 27 \implies S \ge 81$$

Since $81 = 3^4$, the minimum value of our sum is $3^4$. Taking the logarithm, we get $m = \log_3(3^4) = 4$. We have successfully conquered the first peak!

### Phase 2: The Logarithmic Challenge

Now, let us turn our attention to capital $M$, the maximum value of $\log_3 x_1 + \log_3 x_2 + \log_3 x_3$, subject to $x_1 + x_2 + x_3 = 9$. Using the logarithmic property $\log_a p + \log_a q = \log_a(pq)$, we can condense this sum into a single term: $\log_3(x_1 x_2 x_3)$.

To maximize this, we must maximize the product $P = x_1 x_2 x_3$. Once again, we invoke the AM-GM inequality on $x_1, x_2, x_3$:

$$\frac{x_1 + x_2 + x_3}{3} \ge \sqrt[3]{x_1 x_2 x_3}$$

Substituting the sum $x_1 + x_2 + x_3 = 9$, we get $\frac{9}{3} \ge \sqrt[3]{P}$, which simplifies to $3 \ge \sqrt[3]{P}$. Cubing both sides, we find $P \le 27$.

Thus, the maximum value of the product is $27$. Consequently, $M = \log_3(27) = \log_3(3^3) = 3$. The second peak is ours!

### Phase 3: The Grand Finale

We have our two values: $m = 4$ and $M = 3$. The final task is to evaluate the expression $E = \log_2(m^3) + \log_3(M^2)$.

Substituting our findings, we get:

$$E = \log_2(4^3) + \log_3(3^2)$$

$$E = \log_2(64) + \log_3(9)$$

Since $64 = 2^6$ and $9 = 3^2$, the expression simplifies to $6 + 2 = 8$.

The journey concludes with the number **8**. Remember, in mathematics, as in life, the most complex problems often yield to the most elegant principles. Keep practicing, keep visualizing, and keep falling in love with the process!

Enter Numerical Value:

Visualized Solution

Objective and Core Concept

Setup for $m$

Applying AM-GM for $m$

Simplifying the GM

Minimum of the Sum

Finding $m$

Setup for $M$

Applying AM-GM for $M$

Maximum of the Product

Finding $M$

Final Expression Setup

Final Calculation

The Sigma Insight: Relation Between A.M., G.M., and H.M.

The Symphony of Inequalities

Phase 1

Phase 2

Phase 3

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Enter Numerical Value:

Visualized Solution

Objective and Core Concept

Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Applying AM-GM for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Simplifying the GM

Minimum of the Sum

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Applying AM-GM for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M