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đầu tiên ta chứng minh từ đó nhân chéo ⇔2x−2+2x≥6(x2+2x+4)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇔2x−2−−−−−√+2x≥6(x2+2x+4)−−−−−−−−−−−−√x−2+2xx−2√+x2≥3(x2+2x+4)2⇔2x−2+2x≥6(x2+2x+4)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">Bình phg 2 vế (ko âm) đc:$x-2+2x\sqrt{x-2}+x^2\geq \frac{3(x^2+2x+4)}{2}$$\Leftrightarrow .......$ ( rút gọn đi!)$2x\sqrt{x-2}\geq x^2+4x+16$$x^2-2x\sqrt{x-2}+x-2+3x+18\leq 0$$\Leftrightarrow (x-\sqrt{x-2})^2+3x+18\leq 0$ vô lí vì đk!suy ra bpt vô nghiệm!

ĐK:x≥2" role="presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative; background-color: rgb(255, 255, 255);">ĐK:x≥2ĐK:x≥2đầu tiên ta chứng minh 6(x2+2x+4)−2(x+2)>0" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">6(x2+2x+4)−−−−−−−−−−−−√−2(x+2)>06(x2+2x+4)−2(x+2)>0từ đó nhân chéo 2x−2−4≥6(x2+2x+4)−2x−4" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">2x−2−−−−−√−4≥6(x2+2x+4)−−−−−−−−−−−−√−2x−42x−2−4≥6(x2+2x+4)−2x−4⇔2x−2+2x≥6(x2+2x+4)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">⇔2x−2−−−−−√+2x≥6(x2+2x+4)−−−−−−−−−−−−√⇔2x−2+2x≥6(x2+2x+4)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">Bình phg 2 vế (ko âm) đc:$x-2+2x\sqrt{x-2}+x^2\geq \frac{3(x^2+2x+4)}{2}$$\Leftrightarrow $ ( rút gọn đi!)$2x\sqrt{x-2}\geq x^2+4x+16$$$⇔2x−2+2x≥6(x2+2x+4)" role="presentation" style="font-size: 13.696px; display: inline; position: relative;">