Question

Suppose a1, a2. . . , an are n positive real numbers with a1a2 . . . an = 1.
prove that:

(1+a1)*(1+a2)...*(1+an) >= 2^n

I've tried using the arithmetic mean value against the geometric mean value with no success .

Any hint / solution / idea would be greatly appreaciated

Answer 1

Step-by-step explanation:

∵ a₁ , a₂ , a₃ , ....... an are n positive real numbers

∴ we have to use AM ≥ GM

Where AM is Arithmetic mean and GM is Geometric mean.

Now, AM of a₁ , a₂, a₃, a₄, ........., an = (a₁ + a₂ + a₃ + ..... + an)/n

GM of a₁ , a₂, a₃, ........, an = (a₁.a₂.a₃.a₄....an)^(1/n)

∴ (a₁ + a₂ + a₃ + ..... + an)/n ≥ (a₁.a₂.a₃.a₄....an)^(1/n)

∵ a₁.a₂.a₃.a₄......an = 1

So, (a₁ + a₂ + a₃ + ..... + an)/n ≥ 1

⇒a₁ + a₂ + a₃+ ...... + an ≥ n

minimum value of a₁ + a₂ + a₃+ ...... + an = n

Now, take , (1 +a₁), (1 + a₂), (1 + a₃) ....... (1 + an)

AM = {(1 + a₁) + (1 + a₂) + (1 + a₃) + ..... + (1 + an)}/n

GM = {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

Now, AM ≥ GM

{(1 + a₁) + (1 + a₂) + (1 + a₃) + ..... + (1 + an)}/n ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

⇒{(1 + 1 + 1 + 1 ....+ n times) + (a₁ + a₂ + a₃ + .....+ an)}/n ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

⇒(n + n)/n ≥{ (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

⇒2 ≥ {(1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an)}^{1/n}

∴ (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an) ≤ 2ⁿ

Hence, minimum value of (1 + a₁)(1 + a₂)(1 + a₃)(1 + a₄)......(1 + an) = 2ⁿ

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