Question:

<p><strong>Step 1:</strong> Apply Weighted AM-GM Inequality.</p><p>By the weighted AM-GM inequality with weights 1, 2, 3, 4, 5 (which sum to 15):</p><p>$$\frac{1·a + 2·b + 3·c + 4·d + 5·e}{15} \geq \sqrt[15]{a^1 · b^2 · c^3 · d^4 · e^5}$$</p><p><strong>Step 2:</strong> Simplify using given constraints.</p><p>We know $a + b + c + d + e = 15$, so:</p><p>$$\frac{1·a + 2·b + 3·c + 4·d + 5·e}{15} = \frac{a + (a+b+c+d+e) + (b+c+d+e) + (c+d+e) + (d+e) + e}{15}$$</p><p>Actually, using weighted AM-GM directly: since $a + b + c + d + e = 15$ and the weights sum to 15, equality in AM-GM occurs when:</p><p>$$\frac{a}{1} = \frac{b}{2} = \frac{c}{3} = \frac{d}{4} = \frac{e}{5} = k$$</p><p><strong>Step 3:</strong> Find the constant of proportionality.</p><p>From the equality condition: $a = k, b = 2k, c = 3k, d = 4k, e = 5k$</p><p>Substituting into $a + b + c + d + e = 15$:</p><p>$$k + 2k + 3k + 4k + 5k = 15$$</p><p>$$15k = 15$$</p><p>$$k = 1$$</p><p><strong>Step 4:</strong> Verify with the product constraint.</p><p>With $k = 1$: $a = 1, b = 2, c = 3, d = 4, e = 5$</p><p>Check: $ab^2c^3d^4e^5 = 1 · 2^2 · 3^3 · 4^4 · 5^5$</p><p>$= 1 · 4 · 27 · 256 · 3125$</p><p>We need to verify this equals $(120)^3 × 50$:</p><p>$(120)^3 × 50 = 120^3 × 50 = (2^3·3·5)^3 × (2·5^2) = 2^9 · 3^3 · 5^3 × 2 · 5^2 = 2^{10} · 3^3 · 5^5$</p><p>And $1 · 4 · 27 · 256 · 3125 = 2^2 · 3^3 · 2^8 · 5^5 = 2^{10} · 3^3 · 5^5$ ✓</p><p><strong>Step 5:</strong> Calculate the sum of squares.</p><p>$$a^2 + b^2 + c^2 + d^2 + e^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2$$</p><p>$$= 1 + 4 + 9 + 16 + 25 = 55$$</p><p><strong>∴ Answer:</strong> 55</p>