Áp dụng BĐT Svácxơ, ta có:

\(\dfrac{a^2}{b+1}+\dfrac{b^2}{c+1}+\dfrac{c^2}{a+1}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3}=\dfrac{81}{12}=\dfrac{27}{4}\)

Dấu "=" ⇔ a=b=c=3

Áp dụng BĐT Cô-si:

\(\dfrac{a^2}{b+1}+\dfrac{9}{16}\left(b+1\right)\ge2\sqrt{\dfrac{9a^2\left(b+1\right)}{16\left(b+1\right)}}=\dfrac{3a}{2}\) 

Tương tự: \(\dfrac{b^2}{c+1}+\dfrac{9}{16}\left(c+1\right)\ge\dfrac{3b}{2}\) ; \(\dfrac{c^2}{a+1}+\dfrac{9}{16}\left(a+1\right)\ge\dfrac{3c}{2}\)

Cộng vế:

\(VT+\dfrac{9}{16}\left(a+b+c+3\right)\ge\dfrac{3}{2}\left(a+b+c\right)\)

\(\Leftrightarrow VT+\dfrac{27}{4}\ge\dfrac{27}{2}\Rightarrow VT\ge\dfrac{27}{4}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Kiểm tra lại mẫu số của 3 phân thức

Mẫu số của \(b+1\ne c+2,a+2.\)

Xem lại đề bạn

\(\dfrac{a^2}{b+1}+\dfrac{b^2}{c+1}+\dfrac{c^2}{a+1}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3}=\dfrac{9^2}{9+3}=\dfrac{27}{4}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Chứng minh BĐT \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\) với \(\left(a,b,c>0\right)\)

Trước hết ta cm \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)\(\Leftrightarrow\frac{x^2b+y^2a}{ab}\ge\frac{x^2+y^2+2xy}{a+b}\)\(\Leftrightarrow\left(x^2b+y^2a\right)\left(a+b\right)\ge ab\left(x^2+y^2+2xy\right)\)(vì tất cả các tử số và mẫu số đều dương)

\(\Leftrightarrow x^2ab+y^2ab+x^2b^2+y^2a^2\ge abx^2+aby^2+2abxy\)\(\Leftrightarrow x^2b^2-2abxy+y^2a^2\ge0\)\(\Leftrightarrow\left(xb-ya\right)^2\ge0\)(luôn đúng)

Vậy BĐT được cm 

Để có đpcm thì ta chỉ cần áp dụng 2 lần BĐT ta vừa chứng minh xong:

\(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y\right)^2}{a+b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)

Gọi quãng đường AB là x (km)

Thời gian xe máy đi từ A đến B là: x/40(h)

Thời gian xe máy đi từ B về A là: x/50(h)

Đổi 45 phút=3/4 h

Ta có phương trình:

x/40 -x/50 = 3/4

=> 5x - 4x = 150

<=> x = 150

Vậy quãng đường AB dài 150 km

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\(Q=\left(x+\dfrac{2}{x}\right)^2+\left(y+\dfrac{2}{y}\right)^2\ge\dfrac{1}{2}\left(x+\dfrac{2}{x}+y+\dfrac{2}{y}\right)^2\)

\(Q\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(2\sqrt{\dfrac{x}{x}}+2\sqrt{\dfrac{y}{y}}+\dfrac{4}{x+y}\right)^2\)

\(Q\ge\dfrac{1}{2}\left(4+\dfrac{4}{x+y}\right)^2\ge\dfrac{1}{2}\left(4+\dfrac{4}{2}\right)^2=18\)

\(Q_{min}=18\) khi \(x=y=1\)

22+3000

Đề chép sai rồi kìa.

\(\left(x+\frac{2}{x}\right)^2+\left(y+\frac{2}{y}\right)^2=x^2+y^2+\frac{4}{x^2}+\frac{4}{y^2}+4+4\)

\(=\left(x^2+\frac{1}{x^2}\right)+\left(y^2+\frac{1}{y^2}\right)+\left(\frac{3}{x^2}+3x+3x\right)+\left(\frac{3}{y^2}+3y+3y\right)-6\left(x+y\right)+8\)

\(\ge2+2+9+9-6.2+8=18\)

\(C=x^2+y^2+\dfrac{4}{x^2}+\dfrac{4}{y^2}\)

\(=\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)+3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

Áp dụng BĐT Cô si cho 2 số dương, ta có:

\(C\ge2\sqrt{x^2.\dfrac{1}{x^2}}+2\sqrt{y^2.\dfrac{1}{y^2}}+3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

\(=4+3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

Áp dụng BĐT Svácxơ, ta có:

\(C\ge4+3.\dfrac{4}{x^2+y^2}=4+\dfrac{12}{x^2+y^2}\) 

\(C\ge4+\dfrac{12}{2}=4+6=10\)\(\left(x^2+y^2\le2\right)\)

Dấu "=" \(\Leftrightarrow x=y=1\)

\(C=\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)+3\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

\(C\ge2\sqrt{\dfrac{x^2}{x^2}}+2\sqrt{\dfrac{y^2}{y^2}}+\dfrac{3}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge4+\dfrac{3}{2}\left(\dfrac{4}{x+y}\right)^2\ge4+\dfrac{3}{2}.\left(\dfrac{4}{2}\right)^2=10\)

\(C_{min}=10\) khi \(x=y=1\)

\(\dfrac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}=1\)

Áp dụng BĐT Bunhiacopxki:

\(\left(\dfrac{2}{3}+1+\dfrac{1}{3}\right)\left(\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}\right)\ge\left(\sqrt{\dfrac{2}{3}.\dfrac{3}{2}x^2}+\sqrt{1.\left(y+\dfrac{z}{2}\right)^2}+\sqrt{\dfrac{1}{3}.\dfrac{3z^2}{4}}\right)^2\)

\(\Leftrightarrow2.1\ge\left(x+y+\dfrac{z}{2}+\dfrac{z}{2}\right)^2=\left(x+y+z\right)^2\)

\(\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

\(\frac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)

\(\Rightarrow\left(x+y+z\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)