Let $x,y,a,b~\text{and}~ c\in \mathbb{R}$ such that $a,b>0$. Then show that $ax^2+by^2+2cxy \geq 0$ and $(ab-c^2)>0$. Please help to prove this. I tried AM GM inequality but did not succeeded.
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You can always choose $c \in \Bbb R$ such that the expression becomes negative (unless $x=y=0$). Did you forget some condition on $a,b,c$? – Martin R Sep 30 '19 at 09:02
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See https://math.stackexchange.com/q/2193561/42969 for a correct version. – Martin R Sep 30 '19 at 09:05
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Yes, I forgot but now I have mentioned that – MANI Sep 30 '19 at 09:07
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2Possible duplicate of Given $ax^2+by^2+cxy > 0$, what can I deduce about $a$, $b$, and $c$? – Martin R Sep 30 '19 at 09:10
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Also: https://math.stackexchange.com/q/2193561/42969 – Martin R Sep 30 '19 at 09:10
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Sorry, I was not able to find out this question on this site thats why I posted it here. – MANI Sep 30 '19 at 09:11
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1Should I delete this question? Please guide me. – MANI Sep 30 '19 at 09:14
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False. Take $a=b=x=y=1$ and $c =-2$.
Answer for the modified question: $ax^{2}+by^{2}+2cxy \geq ax^{2}+by^{2}-2|c||x||y|=(\sqrt{a}|x|-\sqrt {b}|y|)^{2} +2\sqrt {|a||b|} |x||y| -2c |x||y| \geq 0$.
Kavi Rama Murthy
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1Please guide me what should I do to? It is mentioned above that this is a duplicate question. Will this harm to my reputation on this site? – MANI Sep 30 '19 at 09:18