[Kamal Kamra]

Now we know that $a,b,c$ are positive real numbers and $abc=1$.
$$\frac{a^2+b^2+c^2}{3} \geq {(a^2b^2c^2)}^\frac{1}{3} \space \space \space \space \space (By\space AM\geq GM)) $$
$$\frac{a+b+c}{3} \geq {(abc)}^\frac{1}{3} \space \space \space \space \space (By\space AM\geq GM))$$
Now subtracting both inequalities give:
$$a(a-1)+b(b-1)+c(c-1)\geq {(a^2b^2c^2)}^\frac{1}{3}-{(abc)}^\frac{1}{3}$$
Now because $abc=1$ and $a,b,c$ are positive real numbers:
$$a(a-1)+b(b-1)+c(c-1)\geq 0$$
QED

[Kamal Kamra]

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[Kamal Kamra]

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[Kamal Kamra]

Also please verify this proof and provide an alternate proof if present.

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