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áp dụng bđt bunhiacopxki ta có $\frac{25a}{b+c}+25+\frac{16b}{a+c}+16+\frac{c}{a+b}+1=(a+b+c)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})$$=\frac{1}{2}(b+c+a+c+a+b)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})$$\geq \frac{(5+4+1)^{2}}{2}=50$$\Rightarrow \frac{25a}{b+c}+ \frac{16b}{a+c}+ \frac{c}{a+b} \geq 8$ ta thấy dấu "=" k xảy ra $\Rightarrow$ đpcm

áp dụng bđt bunhiacopxki ta có $\frac{25a}{b+c}+25+\frac{16b}{a+c}+16+\frac{c}{a+b}+1=(a+b+c)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})$$=\frac{1}{2}(b+c+a+c+a+b)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})$$\geq \frac{(5+4+1)^{2}}{2}=50$$\Rightarrow \frac{25a}{b+c}+ \frac{16b}{a+c}+ \frac{c}{a+b} \geq 8$ ta thấy dấu "=" k xảy ra $\Rightarrow$ đpcm

áp dụng bđt bunhiacopxki ta có $\frac{25a}{b+c}+25+\frac{16b}{a+c}+16+\frac{c}{a+b}+1=(a+b+c)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})$$=\frac{1}{2}(b+c+a+c+a+b)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})$$\geq \frac{(5+4+1)^{2}}{2}=50$$\Rightarrow \frac{25a}{b+c}+\frac{16b}{a+c}+\frac{c}{a+b}\geq8$ ta thấy dấu "=" k xảy ra $\Rightarrow$ đpcm

áp dụng bđt bunhiacopxki ta có $\frac{25a}{b+c}+25+\frac{16b}{a+c}+16+\frac{c}{a+b}+1=(a+b+c)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})$$=\frac{1}{2}(b+c+a+c+a+b)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})$$\geq \frac{(5+4+1)^{2}}{2}=50$$\Rightarrow \frac{25a}{b+c}+ \frac{16b}{a+c}+ \frac{c}{a+b} \geq 8$ta thấy dấu "=" k xảy ra$\Rightarrow$ đpcm