Quay lại đáp án

	5	thêm 23 ký tự trong đáp án
TA CÓ : $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$ TA CÓ : $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$ TA CÓ : $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$

	4	bớt 6 ký tự trong đáp án
TA CÓ : $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ ( $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$ TA CÓ : $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ ( $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$ TA CÓ : $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ ( $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$

	3	sửa đáp án
TA CÓ : BĐT $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ (BĐT $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$ TA CÓ : BĐT $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ (BĐT $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$ TA CÓ : BĐT $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ (BĐT $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu = xảy ra $\Leftrightarrow a=b=c$

	2	sửa đáp án
TA CÓ : BĐT $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ (BĐT $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu $=$ xảy ra $\Leftrightarrow a=b=c$ TA CÓ : BĐT $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ (BĐT $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu $=$ xảy ra $\Leftrightarrow a=b=c$ TA CÓ : BĐT $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ (BĐT $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu $=$ xảy ra $\Leftrightarrow a=b=c$

	1	tạo mới
TA CÓ : BĐT $\Leftrightarrow 16\Sigma \frac{1}{2a+b+c}\leq 4(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $\frac{16}{2a+b+c}=\frac{(1+1+1+1)^2}{a+a+b+c}\leq \frac{2}{a}+\frac{1}{b}+\frac{1}{c}$ (BĐT $B-C-S$ )Tương tự như vâỵ ta có đpcm dấu $=$ xảy ra $\Leftrightarrow a=b=c$

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