Câu hỏi của Thức Vương

Question

Cho các số thực dương a,b,c thỏa mãn $2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=6$Tìm MIN: \(P=\frac{bc}{a\left(2b+c\right)}+\frac{ca}{b\left(2a+c\right)}+\frac...

Kiệt Nguyễn · Accepted Answer

$P=\frac{bc}{2ab+ac}+\frac{ca}{2ab+bc}+\frac{4ab}{bc+ca}$Xét $Q=P+3=\frac{bc}{2ab+ac}+1+\frac{ca}{2ab+bc}+1+\frac{4ab}{bc+ca}+1$$Q=\frac{2ab+ac+bc}{2ab+ac}+\frac{2ab+ac+bc}{2ab+bc}+\frac{4ab+bc+ca}{bc+ca}$$=\left(2ab+ac+bc\right)\left(\frac{1}{2ab+ac}+\frac{1}{2ab+bc}\right)+\frac{4ab+bc+ca}{bc+ca}$$\ge\left(2ab+ac+bc\right)\frac{4}{4ab+ac+bc}+\frac{4ab+bc+ca}{bc+ca}=K$(Áp dụng BĐT $\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}$với a, b không âm)$K=\frac{2\left(4ab+ac+bc\right)+2\left(ac+bc\right)}{4ab+ac+bc}+\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}$$+\frac{7\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}$$=2+\left[\frac{2\left(ac+bc\right)}{4ab+ac+bc}+\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\right]+\frac{7}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}$$\ge2+2\sqrt{\frac{2\left(ac+bc\right)}{4ab+ac+bc}.\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}}+\frac{7}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}$(Áp dụng BĐT Cô - si cho 2 số không âm)$=\frac{37}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}$Mặt khác: $6=2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=\frac{2\left(a^2+b^2\right)}{ab}+\frac{c\left(a^3+b^3\right)}{a^2b^2}$$=\frac{2\left(a^2+b^2\right)}{ab}+\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}$$\ge\frac{2.2ab}{ab}+\frac{c\left(a+b\right)\left(2ab-ab\right)}{a^2b^2}=4+\frac{ac+bc}{ab}$(theo BĐT $a^2+b^2\ge2ab$)$\Rightarrow\frac{ac+bc}{ab}\le2\Leftrightarrow\frac{ab}{ac+bc}\ge\frac{1}{2}$$\Rightarrow K\ge\frac{37}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\ge\frac{37}{9}+\frac{7}{9}.\frac{4}{2}=\frac{17}{3}$Ta có $Q=P+3\ge K\ge\frac{17}{3}\Rightarrow P\ge\frac{17}{3}-3=\frac{8}{3}$Đẳng thức xảy ra khi $\hept{\begin{cases}2ab+ac=2ab+bc\\frac{2\left(ac+bc\right)}{4ab+ac+bc}=\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\a=b\end{cases}}$$\Leftrightarrow a=b=c$Câu trả lời: $P=\frac{bc}{2ab+ac}+\frac{ca}{2ab+bc}+\frac{4ab}{bc+ca}$Xét $Q=P+3=\frac{bc}{2ab+ac}+1+\frac{ca}{2ab+bc}+1+\frac{4ab}{bc+ca}+1$$Q=\frac{2ab+ac+bc}{2ab+ac}+\frac{2ab+ac+bc}{2ab+bc}+\frac{4ab+bc+ca}{bc+ca}$$=\left(2ab+ac+bc\right)\left(\frac{1}{2ab+ac}+\frac{1}{2ab+bc}\right)+\frac{4ab+bc+ca}{bc+ca}$$\ge\left(2ab+ac+bc\right)\frac{4}{4ab+ac+bc}+\frac{4ab+bc+ca}{bc+ca}=K$(Áp dụng BĐT $\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}$với a, b không âm)$K=\frac{2\left(4ab+ac+bc\right)+2\left(ac+bc\right)}{4ab+ac+bc}+\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}$$+\frac{7\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}$$=2+\left[\frac{2\left(ac+bc\right)}{4ab+ac+bc}+\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\right]+\frac{7}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}$$\ge2+2\sqrt{\frac{2\left(ac+bc\right)}{4ab+ac+bc}.\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}}+\frac{7}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}$(Áp dụng BĐT Cô - si cho 2 số không âm)$=\frac{37}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}$Mặt khác: $6=2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=\frac{2\left(a^2+b^2\right)}{ab}+\frac{c\left(a^3+b^3\right)}{a^2b^2}$$=\frac{2\left(a^2+b^2\right)}{ab}+\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}$$\ge\frac{2.2ab}{ab}+\frac{c\left(a+b\right)\left(2ab-ab\right)}{a^2b^2}=4+\frac{ac+bc}{ab}$(theo BĐT $a^2+b^2\ge2ab$)$\Rightarrow\frac{ac+bc}{ab}\le2\Leftrightarrow\frac{ab}{ac+bc}\ge\frac{1}{2}$$\Rightarrow K\ge\frac{37}{9}+\frac{7}{9}.\frac{4ab}{ac+bc}\ge\frac{37}{9}+\frac{7}{9}.\frac{4}{2}=\frac{17}{3}$Ta có $Q=P+3\ge K\ge\frac{17}{3}\Rightarrow P\ge\frac{17}{3}-3=\frac{8}{3}$Đẳng thức xảy ra khi $\hept{\begin{cases}2ab+ac=2ab+bc\\frac{2\left(ac+bc\right)}{4ab+ac+bc}=\frac{2\left(4ab+bc+ca\right)}{9\left(ac+bc\right)}\a=b\end{cases}}$$\Leftrightarrow a=b=c$

Tran Le Khanh Linh · Answer

Từ $2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=6\Rightarrow6=\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{2\left(a^2+b^2\right)}{ab}$ta có $a^2+b^2\ge2ab\Rightarrow6=\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{2\left(a^2+b^2\right)}{ab}\ge\frac{c\left(a+b\right)}{ab}+4$$\Rightarrow0< \frac{c\left(a+b\right)}{ab}\le2$Lại có $\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}=\frac{\left(bc\right)^2}{abc\left(2b+c\right)}+\frac{\left(ac\right)^2}{abc\left(2a+c\right)}\ge\frac{\left(bc+ac\right)^2}{2abc\left(a+b+c\right)}$$=\frac{\left[c\left(a+b\right)\right]^2}{2abc\left(a+b+c\right)}$và $abc\left(a+b+c\right)=ab\cdot bc+bc\cdot ba+ab\cdot ca\le\frac{\left(ab+bc+ca\right)^2}{3}$$\Rightarrow\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}\ge\frac{3}{2}\left(\frac{c\left(a+b\right)}{ab+bc+ca}\right)^2=\frac{3}{2}\left(\frac{\frac{c\left(a+b\right)}{ab}}{1+\frac{c\left(a+b\right)}{ab}}\right)^2$Đặt $t=\frac{c\left(a+b\right)}{ab}\Rightarrow P\ge\frac{3t^2}{2\left(1+t\right)^2}+\frac{4}{t}\left(0< t\le2\right)$Có $\frac{3t^2}{2\left(1+t\right)^2}+\frac{4}{t}=\left(\frac{3t^2}{\left(1+t\right)^2}+\frac{4}{t}-\frac{8}{3}\right)+\frac{8}{3}=\frac{-7t^2-8t^2+32t+24}{6t\left(1+t\right)^2}+\frac{8}{3}$$=\frac{\left(t-2\right)\left(-7t^2-22t-12\right)}{6t\left(1+t\right)^2}\ge0\forall t\in(0;2]$=> $\frac{\left(t-2\right)\left(-7t^2-22t-12\right)}{6t\left(1+t\right)^2}+\frac{8}{3}\ge\frac{8}{3}\forall t\in(0;2]\frac{1}{2}$Dấu "=" xảy ra <=> t=2 hay a=b=cCâu trả lời: Từ $2\left(\frac{a}{b}+\frac{b}{a}\right)+c\left(\frac{a}{b^2}+\frac{b}{a^2}\right)=6\Rightarrow6=\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{2\left(a^2+b^2\right)}{ab}$ta có $a^2+b^2\ge2ab\Rightarrow6=\frac{c\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2b^2}+\frac{2\left(a^2+b^2\right)}{ab}\ge\frac{c\left(a+b\right)}{ab}+4$$\Rightarrow0< \frac{c\left(a+b\right)}{ab}\le2$Lại có $\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}=\frac{\left(bc\right)^2}{abc\left(2b+c\right)}+\frac{\left(ac\right)^2}{abc\left(2a+c\right)}\ge\frac{\left(bc+ac\right)^2}{2abc\left(a+b+c\right)}$$=\frac{\left[c\left(a+b\right)\right]^2}{2abc\left(a+b+c\right)}$và $abc\left(a+b+c\right)=ab\cdot bc+bc\cdot ba+ab\cdot ca\le\frac{\left(ab+bc+ca\right)^2}{3}$$\Rightarrow\frac{bc}{a\left(2b+c\right)}+\frac{ac}{b\left(2a+c\right)}\ge\frac{3}{2}\left(\frac{c\left(a+b\right)}{ab+bc+ca}\right)^2=\frac{3}{2}\left(\frac{\frac{c\left(a+b\right)}{ab}}{1+\frac{c\left(a+b\right)}{ab}}\right)^2$Đặt $t=\frac{c\left(a+b\right)}{ab}\Rightarrow P\ge\frac{3t^2}{2\left(1+t\right)^2}+\frac{4}{t}\left(0< t\le2\right)$Có $\frac{3t^2}{2\left(1+t\right)^2}+\frac{4}{t}=\left(\frac{3t^2}{\left(1+t\right)^2}+\frac{4}{t}-\frac{8}{3}\right)+\frac{8}{3}=\frac{-7t^2-8t^2+32t+24}{6t\left(1+t\right)^2}+\frac{8}{3}$$=\frac{\left(t-2\right)\left(-7t^2-22t-12\right)}{6t\left(1+t\right)^2}\ge0\forall t\in(0;2]$=> $\frac{\left(t-2\right)\left(-7t^2-22t-12\right)}{6t\left(1+t\right)^2}+\frac{8}{3}\ge\frac{8}{3}\forall t\in(0;2]\frac{1}{2}$Dấu "=" xảy ra <=> t=2 hay a=b=c

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