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Author |
v6mm131
Post time 2015-1-7 17:15
Proof.We wii consider three cases
Case1.$a\geqslant 4$.In this case,we have
$2b^2+3c^2+abc\geqslant 2b^2+2c^2+4bc=2(b+c)^2$.
Therefore,it suffices to prove that
$a^2+2(b+c)^2\geqslant24$
But this inequality follows immediately from Cauchy-Schwarz Inequality
$a^2+2(b+c)^2\geqslant\frac{(a+b+c)^2}{1+\frac{1}{2}}=24$
Case2.$b\geqslant2$.Since
$a^2+3c^2+abc\geqslant a^2+c^2+2ac=(a+c)^2$
it is enough to check the following inequality
$(a+c)^2+2b^2\geqslant24$
By the Cauchy-Schwarz Inequality,we have
$(a+c)^2+2b^2\geqslant\frac{(a+b+c)^2}{1+\frac{1}{2}}=24$
Case3.$a\leqslant 4$ and $ b\leqslant 2$ In this case, we have$(4-a)(2-b)\geqslant0$,i.e.$ab\geqslant2a+4b-8$,
According to this result,we find that
$a^2+2b^2+3c^2+abc\geqslant a^2+2b^2+3c^2+c(2a+4b-8)=(a+c)^2+2(b+c)^2-8c$
On the other hand ,the Cauchy-Schwarz Inequality gives us
$(a+c)^2+2(b+c)^2\geqslant \frac{(a+c+b+c)^2}{1+\frac{1}{2}}=\frac{2}{3}(c+6)^2$
Combining this with the above inequality，we get
$a^2+2b^2+3c^2+abc\geqslant \frac{2}{3}(c+6)^2-8c=\frac{2}{3}c^2+24\geqslant 24$
The proof is completed .It is easy to see that inequality holds if and only if $a=4,b=2,c=0$. |
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kuing
Post time 2015-1-7 14:00
