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Sequences & Series: Non Negative Reals Range

JEE Maths question with a full step-by-step solution.

Question
If a+b+c+d+e=8a+b+c+d+e=8 and a2+b2+c2+d2+e2=16a^2+b^2+c^2+d^2+e^2=16, where a,b,c,d,ea,b,c,d,e are non-negative reals and the range of ee is [ℓ,mn]\left[\ell,\frac{m}{n}\right] with gcd⁡(m,n)=1\gcd(m,n)=1, then ℓ+m+n\ell+m+n is
A2121correct
B2626
C2020
D3131
Solution
Step 1: For a,b,c,da,b,c,d, the quadratic mean is at least the arithmetic mean:
(a+b+c+d4)2≤a2+b2+c2+d24.\left(\frac{a+b+c+d}{4}\right)^2\le\frac{a^2+b^2+c^2+d^2}{4}.
 Step 2: Substitute a+b+c+d=8−ea+b+c+d=8-e and a2+b2+c2+d2=16−e2a^2+b^2+c^2+d^2=16-e^2:
(8−e4)2≤16−e24 ⇒ 64−16e+e2≤64−4e2.\left(\frac{8-e}{4}\right)^2\le\frac{16-e^2}{4}\ \Rightarrow\ 64-16e+e^2\le 64-4e^2.
 Step 3: 5e2−16e≤0⇒0≤e≤1655e^2-16e\le 0\Rightarrow 0\le e\le\dfrac{16}{5}. So ℓ=0, m=16, n=5\ell=0,\ m=16,\ n=5 and ℓ+m+n=21.\ell+m+n=21. Correct answer: (1)
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