[Toán 10] bất đẳng thức

H

huytrandinh

$.4y^{2}+9\geq 12y$
$.3z^{2}+3\geq 6z$
$.x^{4}+2=x^{4}+\frac{1}{2}+\frac{3}{2}\geq x^{2}+\frac{3}{2}$
$=x^{2}+1+\frac{1}{2}\geq 2x+\frac{1}{2}> 2x$
$=>x^{4}+4y^{2}+3z^{2}+14> 2x+12y+6z$
 
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