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Inequality 1(Vasile Cirtoaje)


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.


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