Cho $a,b,c>0$ thỏa mãn : $a+b+c=2019$ . Tìm $Min$ của $P=\sqrt{2a^2+ab+2b^2}+\sqrt{2b^2+bc+2c^2}+\sqrt{2c^2+ca+2a^2}$
Cho $a,b,c>0$ thỏa mãn : $a+b+c=2019$ . Tìm $Min$ của $P=\sqrt{2a^2+ab+2b^2}+\sqrt{2b^2+bc+2c^2}+\sqrt{2c^2+ca+2a^2}$
Ta có:
`P=\sqrt{2a^2+ab+2b^2}+\sqrt{2b^2+bc+2c^2}+\sqrt{2c^2+ca+2a^2}`
`=\sqrt{2a^2+4ab+2b^2-3ab}+\sqrt{2b^2+4bc+2c^2-3bc}+\sqrt{2c^2+4ca+2a^2-3ca}`
`=\sqrt{2(a+b)^2-3ab}+\sqrt{2(b+c)^2-3bc}+\sqrt{2(c+a)^2-3ca}`
Lại có:
`(a-b)^2>=0`
`<=>a^2-2ab+b^2>=0`
`<=>a^2+2ab+b^2>=4ab`
`<=>(a+b)^2>=4ab`
`<=>(a+b)^2/4>=ab`
`<=>-ab>=-(a+b)^2/4`
`<=>-3ab>=-{3(a+b)^2}/4`
Hoàn toàn tương tự:
`-3bc>=-{3(b+c)^2}/4`; `-3ca>=-{3(c+a)^2}/4`
`=>A>=\sqrt{2(a+b)^2-{3(a+b)^2}/4}+\sqrt{2(b+c)^2-{3(b+c)^2}/4}+\sqrt{2(c+a)^2-{3(c+a)^2}/4}`
`=\sqrt{{5(a+b)^2}/4}+\sqrt{{5(b+c)^2}/4}+\sqrt{{5(c+a)^2}/4}`
`={\sqrt5(a+b)}/2+{\sqrt5(b+c)}/2+{\sqrt5(c+a)}/2`
`=\sqrt5/2(a+b+b+c+c+a)=\sqrt5/2 .2(a+b+c)=2019\sqrt5`
Dấu `=` xảy ra `<=>a=b=c=673`
Đáp án: $P_{min}=2019\sqrt{5}⇔a=b=c=673$
Giải thích các bước giải:
Ta có:
`\sqrt{2a^2+ab+2b^2}=\sqrt{(\frac{3}{4}a^2-\frac{3}{2}ab+\frac{3}{4}b^2)+(\frac{5}{4}a^2+\frac{5}{2}ab+\frac{5}{4}b^2)}`
`=\sqrt{\frac{3}{4}(a^2-2ab+b^2)+\frac{5}{4}(a^2+2ab+b^2)}=\sqrt{\frac{3}{4}(a-b)^2+\frac{5}{4}(a+b)^2}`
`≥\sqrt{\frac{5}{4}(a+b)^2}=\frac{(a+b)\sqrt{5}}{2}`
Chứng minh tương tự:
`\sqrt{2b^2+bc+2c^2}≥\frac{(b+c)\sqrt{5}}{2}`
`\sqrt{2c^2+ca+2a^2}≥\frac{(a+c)\sqrt{5}}{2}`
`⇒P=\sqrt{2a^2+ab+2b^2}+\sqrt{2b^2+bc+2c^2}+\sqrt{2c^2+ca+2a^2}`
`≥\frac{(a+b)\sqrt{5}}{2}+\frac{(b+c)\sqrt{5}}{2}+\frac{(a+c)\sqrt{5}}{2}`
`=(a+b+c)\sqrt{5}=2019\sqrt{5}`
Dấu bằng xảy ra
$\begin{cases}a-b=0\\b-c=0\\c-a=0\\a+b+c=2019\end{cases}⇔a=b=c=673$