Quay lại đáp án

	4	bớt 9 ký tự trong đáp án		Sửa 02-06-16 01:47 AM tran85295 1
$bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$()Giả sử $a\ge b$ .Đặt $a=xb(x \ge 1)$()$\Leftrightarrow [(x^2+x+1)(x+1)]^2 \ge 36x^3 $$VT \ge(3x)^2.4x^2 \ge VP$(ok)Vậy $k=8$ là số cần tìm $bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$().Đặt $a=xb$()$\Leftrightarrow ((x^2+x+1)(x+1))^2 \ge 36x^3\Leftrightarrow (x-1)^2(x^4+6x^3+19x^2+6x+1) \ge0$ (luôn đúng)Vậy $k=8$là giá trị cần tìm $bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$()Giả sử $a\ge b$ .Đặt $a=xb(x \ge 1)$()$\Leftrightarrow [(x^2+x+1)(x+1)]^2 \ge 36x^3 $$VT \ge(3x)^2.4x^2 \ge VP$(ok)Vậy $k=8$ là số cần tìm

	3	bớt 27 ký tự trong đáp án		Sửa 02-06-16 01:42 AM ๖ۣۜPXM๖ۣۜMinh4212♓ 1
$bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$().Đặt $a=xb$()$\Leftrightarrow ((x^2+x+1)(x+1))^2 \ge 36x^3\Leftrightarrow (x-1)^2(x^4+6x^3+19x^2+6x+1) \ge0$ (luôn đúng)Vậy $k=8$là giá trị cần tìm $bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$()Giả sử $a\ge b$ .Đặt $a=xb(x \ge 1)$()$\Leftrightarrow ((x^2+x+1)(x+1))^2 \ge 36x^3\Leftrightarrow (x-1)^2(x^4+6x^3+19x^2+6x+1) \ge0$ (luôn đúng)Vậy $k=8$là giá trị cần tìm $bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$().Đặt $a=xb$()$\Leftrightarrow ((x^2+x+1)(x+1))^2 \ge 36x^3\Leftrightarrow (x-1)^2(x^4+6x^3+19x^2+6x+1) \ge0$ (luôn đúng)Vậy $k=8$là giá trị cần tìm

	2	nhầm 1chút thôi :v		Sửa 02-06-16 01:41 AM ๖ۣۜPXM๖ۣۜMinh4212♓ 1
$bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$()Giả sử $a\ge b$ .Đặt $a=xb(x \ge 1)$()$\Leftrightarrow ((x^2+x+1)(x+1))^2 \ge 36x^3\Leftrightarrow (x-1)^2(x^4+6x^3+19x^2+6x+1) \ge0$ (luôn đúng)Vậy $k=8$là giá trị cần tìm $bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3+b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$()Giả sử $a\ge b$ .Đặt $a=xb(x \ge 1)$()$\Leftrightarrow (x^2+x+1)(x+1) \ge 6\Leftrightarrow (x-1)(x^2+3x+5) \ge0$ (luôn đúng)Vậy $k=8$là giá trị cần tìm $bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$()Giả sử $a\ge b$ .Đặt $a=xb(x \ge 1)$()$\Leftrightarrow ((x^2+x+1)(x+1))^2 \ge 36x^3\Leftrightarrow (x-1)^2(x^4+6x^3+19x^2+6x+1) \ge0$ (luôn đúng)Vậy $k=8$là giá trị cần tìm

	1	tạo mới		Trả lời 02-06-16 01:24 AM tran85295 1
$bdt\Leftrightarrow \frac{1}{a^3}+\frac{1}{b^3}-\frac{4}{a^3+b^3}\ge \frac{16}{(a+b)^3}-\frac{4}{a^3+b^3}+\frac{4k}{(a+b)^3}-\frac{k}{a^3+b^3}$$\Leftrightarrow \frac{(a^3+b^3)^2-4a^3b^3}{(a^3+b^3)a^3b^3} \ge (k+4)\left(\frac{4(a^3+b^3)-(a+b)^3}{(a+b)^3(a^3+b^3)} \right)$$\Leftrightarrow \frac{(a^3-b)^3}{a^3b^3} \ge(k+4)\left(\frac{3(a+b)(a-b)^2}{(a+b)^3} \right)$$\Leftrightarrow (a-b)^2\left[ \frac{(a^2+ab+b^2)^2}{a^3b^3} -\frac{3(k+4)}{(a+b)^2}\right] \ge0$$\Leftrightarrow \frac{(a^2+ab+b^2)^2(a+b)^2}{a^3+b^3} \ge 3(k+4)$()Đk cần : cho $a=b$ thì $VT=36\Leftrightarrow k\le8$Đk đủ, ta sẽ cm $k=8$ là hằng số tốt nhấtThật vậy ()$\Leftrightarrow [(a^2+ab+b^2)(a+b)]^2 \ge 36a^3b^3$()Giả sử $a\ge b$ .Đặt $a=xb(x \ge 1)$()$\Leftrightarrow (x^2+x+1)(x+1) \ge 6\Leftrightarrow (x-1)(x^2+3x+5) \ge0$ (luôn đúng)Vậy $k=8$là giá trị cần tìm

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