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mình k nhầm. câu này yêu cầu cm chứ đâu phải là tìm a, b
Câu 3/ \(\sqrt{\left(x+z\right)^2+\left(y-t\right)^2}+\sqrt{\left(x-z\right)^2+\left(y+t\right)^2}\)
\(\le\sqrt{1+2xz-2yt}+\sqrt{1-2xz+2yt}\)
\(\le\dfrac{1+1+2xz-2yt}{2}+\dfrac{1+1-2xz+2yt}{2}=1+1=2\)
Bài 2:
\(hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+2x\right)+\left(y^2+2y\right)=6\\\left(x^2+2x\right)\left(y^2+2y\right)=9\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+2x=a\\y^2+2y=b\end{matrix}\right.\) thì:\(\left\{{}\begin{matrix}a+b=6\\ab=9\end{matrix}\right.\)
Từ \(a+b=6\Rightarrow a=6-b\) thay vào \(ab=9\)
\(b\left(6-b\right)=9\Rightarrow-b^2+6b-9=0\)
\(\Rightarrow-\left(b-3\right)^2=0\Rightarrow b-3=0\Rightarrow b=3\)
Lại có: \(a=6-b=6-3=3\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+2x=3\\y^2+2y=3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x+3\right)=0\\\left(y-1\right)\left(y+3\right)=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\\left[{}\begin{matrix}y=1\\y=-3\end{matrix}\right.\end{matrix}\right.\)
Bài 3:
\(BDT\Leftrightarrow\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(c+a\right)}+\dfrac{1}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{1}{a^2\left(b+c\right)}\cdot\dfrac{b+c}{4}}\)\(=2\sqrt{\dfrac{1}{4a^2}}=\dfrac{1}{a}\)
Tương tự cho 2 BĐT còn lại ta có:
\(\dfrac{1}{b^2\left(c+a\right)}+\dfrac{c+a}{4}\ge\dfrac{1}{b};\dfrac{1}{c^2\left(a+b\right)}+\dfrac{a+b}{4}\ge\dfrac{1}{c}\)
Cộng theo vế 3 BĐT trên ta có:
\(\Rightarrow VT+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow VT+\dfrac{a+b+c}{2}\ge\dfrac{9}{a+b+c}\ge\dfrac{9}{3\sqrt[3]{abc}}\)
\(\Rightarrow VT+\dfrac{3\sqrt[3]{abc}}{2}\ge\dfrac{9}{3\sqrt[3]{abc}}\Rightarrow VT+\dfrac{3}{2}\ge3\left(abc=1\right)\)
\(\Rightarrow VT\ge\dfrac{3}{2}\). Tức là \(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(c+a\right)}+\dfrac{1}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(a=b=c=1\)
Làm cho hoàn thiện luôn nè
1)ĐK:x>0
pt trở thành: x2+1+3x\(\sqrt{\dfrac{x^2+1}{x}}\)=10x
<=>\(\dfrac{x^2+1}{x}\)+3\(\sqrt{\dfrac{x^2+1}{x}}\)=10(*)
đặt y=\(\sqrt{\dfrac{x^2+1}{x}}\)(y>0)
(*)<=>y2+3y-10=0
<=>(y+5)(y-2)=0
<=>\(\left[{}\begin{matrix}y=-5\\y=2\end{matrix}\right.\)
vậy y =2(y>0)
<=>\(\sqrt{\dfrac{x^2+1}{x}}\)=2<=>x2+1=4x
<=>x2-4x+1=0<=>\(\left[{}\begin{matrix}x=\sqrt{3}+2\\x=2-\sqrt{3}\end{matrix}\right.\)
3) điều phải cm<=>\(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(a+c\right)}+\dfrac{1}{c^2\left(a+b\right)}\ge\dfrac{3}{2}\)đặt x=\(\dfrac{1}{a}\);y=\(\dfrac{1}{b}\);z=\(\dfrac{1}{c}\)
P<=>\(\dfrac{x^2yz}{y+z}+\dfrac{xy^2z}{x+z}+\dfrac{xyz^2}{x+y}\)
=\(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)(xyz=1)
đến đây ta có bất đẳng thức quen thuộc trên
A=\(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)
A+3=\(\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{x+z}+\dfrac{x+y+z}{x+y}\)
=(x+y+z)(\(\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}\))(**)
đặt m=x+y;n=y+z;p=x+z
(**)<=>\(\dfrac{m+n+p}{2}\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)\ge\dfrac{9}{2}\)(điều suy ra được từ bất đẳng thức cô-si cho 3 số)
=>A\(\ge\)\(\dfrac{3}{2}\)
=>P\(\ge\)\(\dfrac{3}{2}\)
ko biết mk làm có đúng ko nhma có gì sai thì đừng trách mk nhé
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{63}{a^2+b^2+c^2}\)
\(6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{a}{ac}\right)+2021\ge\dfrac{54}{ab+bc+ac}+2021\ge\dfrac{54}{a^2+b^2+c^2}+2021\)
<=>\(\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{2021}{9}\)
\(p^2=\left(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\right)^2\)
áp dụng bđt \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
\(p^2\le3.\left(\dfrac{1}{3\left(2a^2+b^2\right)}+\dfrac{1}{3\left(2b^2+c^2\right)}+\dfrac{1}{3\left(2c^2+a^2\right)}\right)=\dfrac{1}{2a^2+b^2}+\dfrac{1}{2b^2+c^2}+\dfrac{1}{2c^2+a^2}\)
\(< =>p^2\le\dfrac{9}{2a^2+b^2+2b^2+c^2+2c^2+a^2}\)
<=> \(p^2\le3.\dfrac{1}{a^2+b^2+c^2}=\dfrac{2021}{3}< =>p\le\sqrt{\dfrac{2021}{3}}\)
dấu bằng xảy ra khi \(a=b=c=\sqrt{\dfrac{3}{2021}}\)
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2021\)
\(\Rightarrow2021\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\sqrt{2021.3}=\sqrt{6063}\)
Từ đó:
\(\sqrt{3\left(2a^2+b\right)}=\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}\ge\sqrt{\left(2a+b\right)^2}=2a+b\)
\(\Rightarrow\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\dfrac{1}{2a+b}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)
Tương tự: \(\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(\Rightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{\sqrt{6063}}{3}\)
\(P_{max}=\dfrac{\sqrt{6063}}{3}\) khi \(a=b=c=\dfrac{3}{\sqrt{6063}}\)
Làm biếng làm quá ký hiện \(\sum\) cho mau nhé
\(\sum\dfrac{\left(a+1\right)^6}{b^5}\ge\sum\dfrac{\left(2\sqrt{a}\right)^6}{b^5}=\sum\dfrac{64a^3}{b^5}\ge64.2\sqrt{\dfrac{1}{a^2b^2}}\)
\(\ge\dfrac{128}{\dfrac{\left(a+b\right)^2}{4}}\ge\dfrac{128}{\dfrac{2^2}{4}}=128\)
phiền ông nha nhưng mà tỉnh t chưa học xích ma nhờ ông ghi chi tít jup t nha tks