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# Inequality 1(Vasile Cirtoaje)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers such that $\displaystyle a^2+b^2+c^2+d^2=1$. Prove that

$\displaystyle (1-a)(1-b)(1-c)(1-d)\geq abcd$

1st solution:

We divide the inequality with $\displaystyle a,b,c,d$. Then we get that $\displaystyle \frac{1-a}{a}\cdot \frac{1-b}{b}\cdot \frac{1-c}{c}\cdot \frac{1-d}{d}\geq 1$. Let $\displaystyle x=\frac{1-a}{a}, y=\frac{1-b}{b}, z=\frac{1-c}{c}, w=\frac{1-d}{d}$.

We need to prove that

$\displaystyle xyzw\geq 1$.

But from the hypothesis we get that $\displaystyle 1=\sum_{cyc}\frac{1}{(1+x)^2}$.

From Jensen’s inequality we get that

$\displaystyle 1=\sum_{cyc}\frac{1}{(1+x)^2}\geq \frac{1}{1+xy}+\frac{1}{1+zw}$.

After some calculations, we get the desired result, that is

$\displaystyle xyzw\geq 1$.

2nd solution:

From AM-GM inequality we get that

$\displaystyle c^2+d^2\geq 2cd\Longrightarrow 1-a^2-b^2\geq 2cd$.

And hence

$\displaystyle 2(1-a)(1-b)-2cd\geq 2(1-a)(1-b)-1+a^2+b^2=(1-a-b)^2\geq 0$

Similarly we can prove that

$\displaystyle (1-c)(1-d)\geq ab$.

So we prooved that $\displaystyle (1-a)(1-b)\geq cd$. Similarly we can prove that $\displaystyle (1-c)(1-d)\geq ab$. Multiplying these inequalities the desired inequality follows, Q.E.D.