1)cho abcd =1. CMR:
[TEX]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+ \frac{9}{a+b+c+d}\geq \frac{25}{4}[/TEX]
2) cho a,b,c,d >0. CMR:
[TEX]\frac{{a}^{4}}{a{b}^{2}+1}+\frac{{b}^{4}}{b{c}^{2}+1}+\frac{{c}^{4}}{c{a}^{2}+1}\geq \frac{abc(a+b+c)}{1+abc}[/TEX]
Bài 1
dùng dồn biến SMV chẳng hạn, giả sử
[TEX]\begin{array}{l}a \ge b \ge c \ge d = > bd \le 1\\f\left( {a,b,c,d} \right) - f\left( {a,\sqrt {bd} ,c,\sqrt {bd} } \right) = \left( {\frac{1}{{bd}} - \frac{9}{{\left( {a + b + c + d} \right)\left( {a + 2\sqrt {bd} + c} \right)}}} \right){\left( {\sqrt b - \sqrt d } \right)^2}\\\frac{1}{{bd}} \ge 1;\frac{9}{{\left( {a + b + c + d} \right)\left( {a + 2\sqrt {bd} + c} \right)}} \le \frac{9}{{16}} = > f\left( {a,b,c,d} \right) - f\left( {a,\sqrt {bd} ,c,\sqrt {bd} } \right) \ge 0\end{array}[/TEX]
do đó ta chỉ cần chứng minh bđt khi b=c=d hay là
[TEX]\begin{array}{l}a{t^3} = 1\\\frac{1}{a} + \frac{3}{t} + \frac{9}{{a + 3t}} \ge \frac{{25}}{4}\\ \Leftrightarrow {t^3} + \frac{3}{t} + \frac{9}{{\frac{1}{{{t^3}}} + 3t}} \ge \frac{{25}}{4} \Leftrightarrow {\left( {t - 1} \right)^2}\left( {12{t^6} + 24{t^5} + 36{t^4} - 27{t^3} - 14{t^2} - t + 12} \right) \ge 0\\12{t^6} + 24{t^5} + 36{t^4} - 27{t^3} - 14{t^2} - t + 12 > {\left( {6{t^2} - 3t - 3} \right)^2} \ge 0\end{array}[/TEX]
suy ra dpcm
bài 2
có
[TEX]VT = \sum\limits_{cyc} {\frac{{{a^4}}}{{a{b^2} + 1}}} \ge \frac{{{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}}{{\sum\limits_{cyc} {a{b^2} + 3} }}[/TEX]
ta sẽ cm
[TEX]\begin{array}{l}\frac{{{{\left( {\sum\limits_{cyc} {{a^2}} } \right)}^2}}}{{\sum\limits_{cyc} {a{b^2} + 3} }} \ge \frac{{abc\left( {a + b + c} \right)}}{{1 + abc}} \Leftrightarrow {\left( {{a^2} + {b^2} + {c^2}} \right)^2} - 3abc\left( {a + b + c} \right) + abc\left[ {{{\left( {{a^2} + {b^2} + {c^2}} \right)}^2} - \left( {a + b + c} \right)\left( {a{b^2} + b{c^2} + c{a^2}} \right)} \right] \ge 0\\{\left( {{a^2} + {b^2} + {c^2}} \right)^2} - 3abc\left( {a + b + c} \right) \ge 0\\{a^4} + {b^4} + {c^4} \ge a{b^3} + b{c^3} + c{a^3}\\{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} \ge abc\left( {a + b + c} \right)\\= > {\left( {{a^2} + {b^2} + {c^2}} \right)^2} - \left( {a + b + c} \right)\left( {a{b^2} + b{c^2} + c{a^2}} \right) \ge 0\end{array}[/TEX]
suy ra dpcm
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