Câu hỏi của Gillgames

Question

Cho a,b,c > 0 thỏa mãn $\left(a+b\right)\left(b+c\right)\left(c+a\right)=1$. Tìm giá trị nhỏ nhất của biểu thức:

\(P=\dfrac{\sqrt{a^2-ab+b^2}}{\sqrt{ab+1}}+\dfrac{\sqrt{b^2-bc+c^2}}{\sqrt{bc+1}}+\d...

Lightning Farron · Accepted Answer

máy lag + mệt = nản, vô đây tham khảo HERECâu trả lời: máy lag + mệt = nản, vô đây tham khảo HERE

Neet · Answer

ta có :$a^2-ab+b^2=\left(a+b\right)^2-3ab\ge\left(a+b\right)^2-\dfrac{3}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2$(theo BĐT AM-GM)

$\Rightarrow P\ge\sum\dfrac{a+b}{2\sqrt{ab+1}}$

ÁP dụng BĐT AM-GM:

$\dfrac{a+b}{2\sqrt{ab+1}}+\dfrac{b+c}{2\sqrt{bc+1}}+\dfrac{c+a}{2\sqrt{ca+1}}\ge3\sqrt[3]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}=\dfrac{3}{2}.\dfrac{1}{\sqrt[3]{\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}$

Mà $\sqrt[3]{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\le\dfrac{1}{3}\left(ab+bc+ca+3\right)$

$\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\left(ab+bc+ca+3\right)}}$(*)

ta liên tưởng đến BĐT phụ:$\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)$

Cm: phân tích :$VT=xy\left(x+y\right)+yz\left(y+z\right)+zx\left(x+z\right)+2xyz$

$=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(z+x\right)+3xyz-xyz$

$=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz$

mà $\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz$

nên $\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)$

Áp dụng:

$1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)$

mặt khác,theo AM-GM,dễ dàng chứng minh được $a+b+c\ge\dfrac{3}{2}$

nên $1\ge\dfrac{8}{9}.\dfrac{3}{2}\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le\dfrac{3}{4}$

từ (*)$\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\dfrac{3}{4}+3}}=\dfrac{3}{\sqrt{5}}$

Dấu = xảy ra khi $a=b=c=\dfrac{1}{2}$Câu trả lời: ta có :$a^2-ab+b^2=\left(a+b\right)^2-3ab\ge\left(a+b\right)^2-\dfrac{3}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2$(theo BĐT AM-GM)