MathLabs

Problem 3

Let nn be a positive integer and a1,a2,…,ana_1,a_2,\ldots,a_n be positive real numbers. Prove that ∑i=1n12i(21+ai)2i≥21+a1a2⋯an−12n.\sum_{i=1}^{n}\frac{1}{2^i}\left(\frac{2}{1+a_i}\right)^{2^i}\ge\frac{2}{1+a_1a_2\cdots a_n}-\frac{1}{2^n}.
Step 3 of 5: Inductive step of the lemma
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Ak(x)=2k−2Ak−1(x)2,2(Ak−1(x)2+Ak−1(y)2)≥(Ak−1(x)+Ak−1(y))2A_k(x)=2^{k-2}A_{k-1}(x)^2,\qquad 2\big(A_{k-1}(x)^2+A_{k-1}(y)^2\big)\ge\big(A_{k-1}(x)+A_{k-1}(y)\big)^2
Detailed analysis

For k≥2k\ge2, squaring the definition gives Ak(x)=2k−2Ak−1(x)2A_k(x)=2^{k-2}A_{k-1}(x)^2. Since 2(p2+q2)≥(p+q)22(p^2+q^2)\ge(p+q)^2 for all reals p,qp,q, applying this with p=Ak−1(x)p=A_{k-1}(x), q=Ak−1(y)q=A_{k-1}(y) and then the inductive hypothesis Ak−1(x)+Ak−1(y)≥Ak−2(xy)A_{k-1}(x)+A_{k-1}(y)\ge A_{k-2}(xy) gives Ak(x)+Ak(y)=2k−2(Ak−1(x)2+Ak−1(y)2)≥2k−3(Ak−1(x)+Ak−1(y))2≥2k−3Ak−2(xy)2=Ak−1(xy)A_k(x)+A_k(y)=2^{k-2}\big(A_{k-1}(x)^2+A_{k-1}(y)^2\big)\ge2^{k-3}\big(A_{k-1}(x)+A_{k-1}(y)\big)^2\ge2^{k-3}A_{k-2}(xy)^2=A_{k-1}(xy). Hence Ak(x)+Ak(y)≥Ak−1(xy)A_k(x)+A_k(y)\ge A_{k-1}(xy) holds for every k≥1k\ge1.