How to prove that $b^2c^2+c^2a^2+a^2b^2 \gt abc(a+b+c)$
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HINT: use that $$x^2+y^2+z^2\geq xy+yz+zx$$ after my hint above we have $$a^2b^2+b^2c^2+c^2a^2\geq a^2bc+ab^2c+abc^2$$
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Author by
learner
Updated on August 07, 2020Comments
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learner almost 3 years
I am stuck with the following problem.
If $a>0$, $b>0$, $c >0$ and not all equal then prove that: $$b^2c^2+c^2a^2+a^2b^2 \gt abc(a+b+c).$$
Additional info:I'm looking for solutions using AM-GM .
I don't know how to progress . I will be grateful if someone explains . Thanks in advance ..
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learner almost 6 yearsGot it. Elegant ..Thanks a lot sir.
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Dr. Sonnhard Graubner almost 6 yearsyou are welcome, this inequality is very usefull -
user236182 almost 6 yearsProof: $(x-y)^2+(y-z)^2+(z-x)^2\ge 0$. Or $x^2+y^2\ge 2xy$, etc. (add the inequalities).
