Câu hỏi của Hồ Văn Đạt

Question

cho a,b>0 thỏa mãn a+b=1. chứng minh (a+1/b)^2 + (b+1/a)^2 lớn hơn hoặc bằng 25/2

zZz Cool Kid_new zZz · Accepted Answer

$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2$$\ge\frac{\left(a+b+\frac{1}{a}+\frac{1}{b}\right)^2}{2}$$\ge\frac{\left(a+b+\frac{4}{a+b}\right)^2}{2}$$=\frac{25}{2}$ tại a=b=1/2Câu trả lời: $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2$$\ge\frac{\left(a+b+\frac{1}{a}+\frac{1}{b}\right)^2}{2}$$\ge\frac{\left(a+b+\frac{4}{a+b}\right)^2}{2}$$=\frac{25}{2}$ tại a=b=1/2

Lê Tài Bảo Châu · Answer

thêm ít cáchCách 1:Áp dụng BĐT bunhiacopxki ta được:$\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]\left(1^2+1^2\right)\ge\left[\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{a}\right)\right]^2$$\Leftrightarrow\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]2\ge\left(1+\frac{1}{a}+\frac{1}{b}\right)^2$(1)Ta có:$\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}$( tự CM nha )ÁP dụng BĐT AM-GM ta có:$\sqrt{ab}\le\frac{a+b}{2}=\frac{1}{2}$$\Rightarrow\frac{1}{a}+\frac{1}{b}\ge4$(2)Thay (2) vào (1) ta được: $\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]2\ge25$$\Rightarrow\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\ge\frac{25}{2}\left(đpcm\right)$Dấu"="xảy ra $\Leftrightarrow a=b=\frac{1}{2}$Cách 2: Đặt $P=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2$Ta có: $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2=a^2+\frac{2a}{b}+\frac{1}{b^2}+b^2+\frac{2b}{a}+\frac{1}{a^2}$$=a^2+\frac{2a}{b}+\frac{1}{16b^2}+\frac{15}{16b^2}+b^2+\frac{2b}{a}+\frac{1}{16a^2}+\frac{15}{16a^2}$$=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\left(\frac{2a}{b}+\frac{2b}{a}\right)+\left(\frac{15}{16b^2}+\frac{15}{16a^2}\right)$ÁP dụng BĐT AM-GM ta có:$a^2+\frac{1}{16a^2}\ge2\sqrt{a^2.\frac{1}{16a^2}}\ge\frac{1}{2}$(3)$b^2+\frac{1}{16b^2}\ge2\sqrt{b^2.\frac{1}{16b^2}}\ge\frac{1}{2}$(4)$\frac{2a}{b}+\frac{2b}{a}\ge2\sqrt{\frac{2a}{b}.\frac{2b}{a}}\ge4$(5)$\frac{15}{16a^2}+\frac{15}{16b^2}\ge2\sqrt{\frac{15.15}{16.16a^2b^2}}=\frac{15}{8ab}$(1) ÁP dụng BĐT AM-GM ta có:$ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}$(2)Thay (2) vào (1) ta được:$\frac{15}{16a^2}+\frac{15}{16b^2}\ge\frac{15}{2}$(6)Cộng (3)+(4)+(5)+(6) ta được: $P\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{2}+4=\frac{25}{2}$Dấu"="xảy ra $\Leftrightarrow a=b=\frac{1}{2}$Cách 3:Làm tắt thui ạĐặt $P=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2=a^2+\frac{2a}{b}+\frac{1}{b^2}+b^2+\frac{2b}{a}+\frac{1}{a^2}\ge2ab+\frac{2}{ab}+4$$P\ge2\left(ab+\frac{1}{ab}\right)+4$$P\ge2\left(ab+\frac{1}{16ab}+\frac{15}{16ab}\right)+4$giống cách 2 rồi làm nốtCâu trả lời: thêm ít cáchCách 1:Áp dụng BĐT bunhiacopxki ta được:$\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]\left(1^2+1^2\right)\ge\left[\left(a+\frac{1}{b}\right)+\left(b+\frac{1}{a}\right)\right]^2$$\Leftrightarrow\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]2\ge\left(1+\frac{1}{a}+\frac{1}{b}\right)^2$(1)Ta có:$\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}$( tự CM nha )ÁP dụng BĐT AM-GM ta có:$\sqrt{ab}\le\frac{a+b}{2}=\frac{1}{2}$$\Rightarrow\frac{1}{a}+\frac{1}{b}\ge4$(2)Thay (2) vào (1) ta được: $\left[\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\right]2\ge25$$\Rightarrow\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2\ge\frac{25}{2}\left(đpcm\right)$Dấu"="xảy ra $\Leftrightarrow a=b=\frac{1}{2}$Cách 2: Đặt $P=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2$Ta có: $\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2=a^2+\frac{2a}{b}+\frac{1}{b^2}+b^2+\frac{2b}{a}+\frac{1}{a^2}$$=a^2+\frac{2a}{b}+\frac{1}{16b^2}+\frac{15}{16b^2}+b^2+\frac{2b}{a}+\frac{1}{16a^2}+\frac{15}{16a^2}$$=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\left(\frac{2a}{b}+\frac{2b}{a}\right)+\left(\frac{15}{16b^2}+\frac{15}{16a^2}\right)$ÁP dụng BĐT AM-GM ta có:$a^2+\frac{1}{16a^2}\ge2\sqrt{a^2.\frac{1}{16a^2}}\ge\frac{1}{2}$(3)$b^2+\frac{1}{16b^2}\ge2\sqrt{b^2.\frac{1}{16b^2}}\ge\frac{1}{2}$(4)$\frac{2a}{b}+\frac{2b}{a}\ge2\sqrt{\frac{2a}{b}.\frac{2b}{a}}\ge4$(5)$\frac{15}{16a^2}+\frac{15}{16b^2}\ge2\sqrt{\frac{15.15}{16.16a^2b^2}}=\frac{15}{8ab}$(1) ÁP dụng BĐT AM-GM ta có:$ab\le\frac{\left(a+b\right)^2}{4}=\frac{1}{4}$(2)Thay (2) vào (1) ta được:$\frac{15}{16a^2}+\frac{15}{16b^2}\ge\frac{15}{2}$(6)Cộng (3)+(4)+(5)+(6) ta được: $P\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{2}+4=\frac{25}{2}$Dấu"="xảy ra $\Leftrightarrow a=b=\frac{1}{2}$Cách 3:Làm tắt thui ạĐặt $P=\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{a}\right)^2=a^2+\frac{2a}{b}+\frac{1}{b^2}+b^2+\frac{2b}{a}+\frac{1}{a^2}\ge2ab+\frac{2}{ab}+4$$P\ge2\left(ab+\frac{1}{ab}\right)+4$$P\ge2\left(ab+\frac{1}{16ab}+\frac{15}{16ab}\right)+4$giống cách 2 rồi làm nốt

l҉o҉n҉g҉ d҉z҉ · Answer

ÁP DỤNG BẤT ĐẲNG THỨC BUNYAKOVSKY DẠNG PHÂN THỨC TA CÓ : $\left(A+\frac{1}{B}\right)^2+\left(B+\frac{1}{A}\right)^2\ge\frac{\left(A+\frac{1}{B}+B+\frac{1}{A}\right)^2}{2}=\frac{\left(1+\frac{1}{A}+\frac{1}{B}\right)^2}{2}$(1)LẠI CÓ $\frac{1}{A}+\frac{1}{B}\ge\frac{4}{A+B}=\frac{4}{1}=4$(2)TỪ (1) VÀ (2) => $\left(A+\frac{1}{B}\right)^2+\left(B+\frac{1}{A}\right)^2\ge\frac{\left(1+\frac{1}{A}+\frac{1}{B}\right)^2}{2}\ge\frac{\left(1+4\right)^2}{2}=\frac{25}{2}$=> $\left(A+\frac{1}{B}\right)^2+\left(B+\frac{1}{A}\right)^2\ge\frac{25}{2}$(ĐPCM)ĐẲNG THỨC XẢY RA <=> A = B = 1/2Câu trả lời: ÁP DỤNG BẤT ĐẲNG THỨC BUNYAKOVSKY DẠNG PHÂN THỨC TA CÓ : $\left(A+\frac{1}{B}\right)^2+\left(B+\frac{1}{A}\right)^2\ge\frac{\left(A+\frac{1}{B}+B+\frac{1}{A}\right)^2}{2}=\frac{\left(1+\frac{1}{A}+\frac{1}{B}\right)^2}{2}$(1)LẠI CÓ $\frac{1}{A}+\frac{1}{B}\ge\frac{4}{A+B}=\frac{4}{1}=4$(2)TỪ (1) VÀ (2) => $\left(A+\frac{1}{B}\right)^2+\left(B+\frac{1}{A}\right)^2\ge\frac{\left(1+\frac{1}{A}+\frac{1}{B}\right)^2}{2}\ge\frac{\left(1+4\right)^2}{2}=\frac{25}{2}$=> $\left(A+\frac{1}{B}\right)^2+\left(B+\frac{1}{A}\right)^2\ge\frac{25}{2}$(ĐPCM)ĐẲNG THỨC XẢY RA <=> A = B = 1/2

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