Let $a,b,c>0$ with$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that $(a + 1)(b + 1)(c + 1) \geq 64$
Ohk so we are given that $abc=a+b+c$ with that now the inequality becomes $2abc+(a+b+c)+1 \geq 64$
How do i proceed from here?
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1The first condition gives you $ab+bc+ca=abc$ not $abc=a+b+c$. – Anurag A Jun 01 '20 at 16:05
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Another one: https://math.stackexchange.com/q/3026163/42969 – Martin R Jun 01 '20 at 18:04
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$$a+1=1+\frac ab+\frac ac+1$$ etc. Multiply this with the similar expressions for $b+1$ and $c+1$ and use AM/GM.
Angina Seng
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Actually i am a beginner to olympiad number theory and inequalities. Could you please explain further? – Aayam Mathur Jun 01 '20 at 16:22
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See https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means for AM/GM. It is one of the cornerstones of inequalities. – Angina Seng Jun 01 '20 at 16:22
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Put $x=\frac 1a,\,y=\frac 1b,\,z=\frac 1z$ then $x+y+z=1.$ Inequality become $$\left(\frac 1x+1\right)\left(\frac 1y+1\right)\left(\frac 1z+1\right)\geqslant 64,$$ or $$\left (\frac{2x+y+z}{x} \right)\left (\frac{2y+z+x}{y} \right)\left (\frac{2z+x+y}{z} \right) \geqslant 64,$$ or $$(2x+y+z)(2z+x+y)(2z+x+y) \geqslant 64. \quad (1)$$ Suppose $z=\min\{x,y,z\}.$ We can write $(1)$ become $$2(x+y+8z)(x-y)^2+(9x+9y+2z)(x-z)(y-z) \geqslant 0.$$ Which is true. We are done.
Nguyenhuyen_AG
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How is that different from this answer to the possible duplicate target? – Martin R Jun 03 '20 at 07:17
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@MartinR I did not follow Sai Teja idea. I just updated my answer – Nguyenhuyen_AG Jun 03 '20 at 07:33
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Now, after homogenization we need to prove that: $$2(ab+ac+bc)+\frac{(a+b+c)abc}{ab+ac+bc}\geq\frac{63a^2b^2c^2}{(ab+ac+bc)^2}$$ or $$2(ab+ac+bc)^3+(a+b+c)(ab+ac+bc)abc\geq63a^2b^2c^2,$$ which is obvious by AM-GM: $$2(ab+ac+bc)^3+(a+b+c)(ab+ac+bc)abc\geq$$ $$\geq2\left(3\sqrt[3]{a^2b^2c^2}\right)^3+abc\cdot3\sqrt[3]{abc}\cdot3\sqrt[3]{a^2b^2c^2}=63a^2b^2c^2,$$
Michael Rozenberg
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