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Inequalities
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Wednesday, February 22, 2012
if x,y are real number
Prove:
Let
then we obtain:
since
So pq=1 put it into x and y we obtain the identity
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(y+\sqrt{x^2+1})=1\,\,\mathrm{prove\,\,that}\,x+y=0 "(x^2+\sqrt{y^2+1})(y^2+\sqrt{x^2+1})=1\,\,\mathrm{prove\,\,that}\,x+y=0")
,\sqrt{x^2+1}=\frac{1}{2}(p+\frac{1}{p})\,p>0 "\dpi{120} x=\frac{1}{2}(p-\frac{1}{p}),\sqrt{x^2+1}=\frac{1}{2}(p+\frac{1}{p})")
,\sqrt{y^2+1}=\frac{1}{2}(q+\frac{1}{q})\,q>0 "\dpi{120} y=\frac{1}{2}(q-\frac{1}{q}),\sqrt{y^2+1}=\frac{1}{2}(q+\frac{1}{q})")
(y+\sqrt{x^2+1})-1=[\frac{1}{2}(p-\frac{1}{p}+q+\frac{1}{q})][\frac{1}{2}(q-\frac{1}{q}+p+\frac{1}{p})]-1 "\dpi{120} (x+\sqrt{y^2+1})(y+\sqrt{x^2+1})-1=[\frac{1}{2}(p-\frac{1}{p}+q+\frac{1}{q})][\frac{1}{2}(q-\frac{1}{q}+p+\frac{1}{p})]-1")
^2+pq(p+q)^2}{4pq^2}(pq-1)=0 "\dpi{120} =\frac{(p-q)^2+pq(p+q)^2}{4pq^2}(pq-1)=0")
^2+pq(p+q)^2}{4pq^2}>0 "\dpi{120} \frac{(p-q)^2+pq(p+q)^2}{4pq^2}>0")
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