[Toán 10] bất đẳng thức

H

huytrandinh

.4y2+9≥12y.4y^{2}+9\geq 12y
.3z2+3≥6z.3z^{2}+3\geq 6z
.x4+2=x4+12+32≥x2+32.x^{4}+2=x^{4}+\frac{1}{2}+\frac{3}{2}\geq x^{2}+\frac{3}{2}
=x2+1+12≥2x+12>2x=x^{2}+1+\frac{1}{2}\geq 2x+\frac{1}{2}> 2x
=>x4+4y2+3z2+14>2x+12y+6z=>x^{4}+4y^{2}+3z^{2}+14> 2x+12y+6z
 
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