quyhktn-qa1的2个不等式

陈洪葛 posted @ Jul 21, 2013 06:45:59 PM in 不等式 , 972 阅读

Problem 1:Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3.$ Prove that
\[   \sqrt{a^3+2bc}+\sqrt{b^3+2ca}+\sqrt{c^3+2ab} \geq 3\sqrt{3}.\]

Proof:by Holder inequality,we have
\[\left(\sum{\sqrt{a^3+2bc}}\right)^2\cdot\left[\sum\frac{(a^2+2bc)^3}{a^3+2bc}\right]\geq (a+b+c)^6 \]
Therefore,it's suffice to prove
\[\sum\frac{(a^2+2bc)^3}{a^3+2bc}\leq 27 \]
by Holder inequality,we can see that
\[ (a+2bc)(a^3+2bc)(a^2+2bc)\geq (a^2+2bc)^3 \]
which imply
\[ \frac{(a^2+2bc)^3}{a^3+2bc}\leq (a+2bc)(a^2+2bc) \]
Thus,it's enough to prove
\[ \sum{(a+2bc)(a^2+2bc)}\leq 27 \]
After homogenous,it's
\[ (\sum{a})(\sum{a^3}+6abc)+12\sum{a^2b^2}+6\sum{a^2bc}\leq (a+b+c)^4 \]
Or
\[ ab(a-b)^2+bc(b-c)^2+ca(c-a)^2\geq 0 \]
Which is obviously true,Hence we are done! Equality occurs when $a=b=c=1$.    $\square$

 

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Problem 2:      If $a,b,c$ are positive real numbers such that $a+b+c=3,$ then
\[ \dfrac{a}{12a+5b^2}+\dfrac{b}{12b+5c^2}+\dfrac{c}{12c+5a^2} \leq \dfrac{3}{17}.\]

Proof:First,we can rewrite the inequality into
\[ \frac{b^2}{4a^2+5b^2+4ab+4ac}+\frac{c^2}{4b^2+5c^2+4bc+4ba}+\frac{a^2}{4c^2+5a^2+4ca+4bc}\geq \frac{3}{17}\]
by Cauchy-Schwarz inequality,we have
\[\left(\sum_{cyc}{\frac{b^2}{4a^2+5b^2+4ab+4ac}}\right)\left[\sum_{cyc}{(2b+c)^2(4a^2+5b^2+4ab+4ac)}\right]\geq \left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2\]
Therefore,it's suffice to prove that
\[ 17\left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2\geq 3\left[\sum_{cyc}{(2b+c)^2(4a^2+5b^2+4ab+4ac)}\right]\]
after some computation,it's
\[ 8\sum_{cyc}{a^4}+20\sum_{cyc}{ab^3}-4\sum_{cyc}{a^3b}-102\sum{a^2bc}+78\sum{a^2b^2}\geq 0 \]
Now,Using Vo Quoc Ba Can's sum of square technique,for $m=8,n=78,p=20,g=-4$,we have to check
\[ \left\{\begin{array}{ll}
        3m(m+n)\geq p^2+pg+g^2 ,\\
        m>0 ,
     \end{array}
\right.\]
it can be easily see that $ 3m(m+n)=2064>336=p^2+pg+g^2 $,Hence the inequality is holds.
Done!          $\square$
 


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