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Lời giải:
a)
Sử dụng pp biến đổi tương đương:
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}\geq \frac{2}{ab+1}\Leftrightarrow \frac{a^2+b^2+2}{(a^2+1)(b^2+1)}\geq \frac{2}{ab+1}\)
\(\Leftrightarrow (ab+1)(a^2+b^2+2)\geq 2(a^2b^2+a^2+b^2+1)\)
\(\Leftrightarrow ab(a^2+b^2)+2ab\geq 2a^2b^2+a^2+b^2\)
\(\Leftrightarrow ab(a^2+b^2-2ab)-(a^2+b^2-2ab)\geq 0\)
\(\Leftrightarrow ab(a-b)^2-(a-b)^2\geq 0\)
\(\Leftrightarrow (ab-1)(a-b)^2\geq 0\) (luôn đúng với mọi $ab\geq 1$)
Ta có đpcm.
b) Áp dụng công thức của phần a ta có:
\(\frac{1}{a^4+1}+\frac{1}{b^4+1}\geq \frac{2}{1+(ab)^2}\)
Tiếp tục áp dụng công thức phần a: \(\frac{1}{1+(ab)^2}+\frac{1}{1+b^4}\geq \frac{2}{1+ab^3}\)
Do đó:
\(\frac{1}{a^4+1}+\frac{3}{b^4+1}\geq \frac{4}{1+ab^3}\)
Hoàn toàn tương tự: \(\frac{1}{b^4+1}+\frac{3}{c^4+1}\geq \frac{4}{1+bc^3}; \frac{1}{c^4+1}+\frac{3}{a^4+1}\geq \frac{4}{1+ca^3}\)
Cộng theo vế các BĐT trên thu được:
\(4\left(\frac{1}{a^4+1}+\frac{1}{b^4+1}+\frac{1}{c^4+1}\right)\geq 4\left(\frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\right)\)
\(\Leftrightarrow \frac{1}{a^4+1}+\frac{1}{b^4+1}+\frac{1}{c^4+1}\geq \frac{1}{1+ab^3}+\frac{1}{1+bc^3}+\frac{1}{1+ca^3}\)
Ta có đpcm
Dấu bằng xảy ra khi $a=b=c=1$
Áp dụng BĐT cauchy ngược dấu ta có:
\(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)
Chứng minh tương tự ta có:
\(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)
Từ đó ta có: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge1-\dfrac{a}{2}+1-\dfrac{b}{2}+1-\dfrac{c}{2}=\)\(=3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\left(đpcm\right)\)
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\) ≥ \(\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+3}=\dfrac{9}{a^2+b^2+c^2+3}\left(1\right)\)
Ta có BĐT : \(a^2+b^2+c^2\text{≥}ab+bc+ac\)
⇔ \(3\left(a^2+b^2+c^2\right)\text{≥}\left(a+b+c\right)^2\)
⇔ \(a^2+b^2+c^2\text{≥}\dfrac{9}{3}=3\left(2\right)\)
Từ ( 1 ; 2 ) ⇒ đpcm .
"=" ⇔ \(a=b=c=\dfrac{1}{3}\)
Cho a,b,c>0 t/m a+b+c=3
CMR: \(\dfrac{a+1}{b^2+1}\)+\(\dfrac{b+1}{c^2+1}\)+\(\dfrac{c+1}{a^2+1}\)>=3
\(\dfrac{a+1}{b^2+1}=a+1-\dfrac{ab^2+b^2}{b^2+1}\) minh hong biet phai lam gi tiep theo dau
Câu 3/ \(\sqrt{\left(x+z\right)^2+\left(y-t\right)^2}+\sqrt{\left(x-z\right)^2+\left(y+t\right)^2}\)
\(\le\sqrt{1+2xz-2yt}+\sqrt{1-2xz+2yt}\)
\(\le\dfrac{1+1+2xz-2yt}{2}+\dfrac{1+1-2xz+2yt}{2}=1+1=2\)
Đặt \(\left(a,b,c\right)\rightarrow\left(\dfrac{x}{y},\dfrac{y}{z},\dfrac{z}{x}\right)\)
BĐT cần c/m tương đương với
\(\sum\dfrac{yz}{xy+xz+2yz}\le\dfrac{3}{4}\)
\(\Leftrightarrow\sum\dfrac{xy+xz}{xy+xz+2yz}\ge\dfrac{3}{2}\)
Ta có \(\sum\dfrac{xy+xz}{xy+xz+2yz}\ge\dfrac{\left(2\sum xy\right)^2}{\sum\left(xy+xz+2yz\right)\left(xy+xz\right)}=\dfrac{4\left(\sum xy\right)^2}{2\sum x^2y^2+6\sum x^2yz}\)
Như vậy ta cần c/m \(\dfrac{4\left(\sum xy\right)^2}{2\sum x^2y^2+6\sum x^2yz}\ge\dfrac{3}{2}\)
\(\Leftrightarrow8\left(\sum xy\right)^2\ge6\sum x^2y^2+18\sum x^2yz\)
\(\Leftrightarrow8\left(\sum xy\right)^2\ge6\left(\sum xy\right)^2+6\sum x^2yz\)
\(\Leftrightarrow\left(\sum xy\right)^2\ge3\sum x^2yz\) (luôn đúng)
Ta có:
\(\dfrac{1}{ab+a+2}\le\dfrac{1}{4}\left(\dfrac{1}{ab+1}+\dfrac{1}{a+1}\right)=\dfrac{1}{4}\left(\dfrac{c}{1+c}+\dfrac{1}{a+1}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{4}\left(\dfrac{a+1}{a+1}+\dfrac{b+1}{b+1}+\dfrac{c+1}{c+1}\right)=\dfrac{3}{4}\)
\(\left\{{}\begin{matrix}\dfrac{1}{a+2}=\dfrac{1}{2}-\dfrac{1}{b+2}+\dfrac{1}{2}-\dfrac{1}{c+2}=\dfrac{b}{2\left(b+2\right)}+\dfrac{c}{2\left(c+2\right)}\ge\sqrt{\dfrac{bc}{\left(b+2\right)\left(c+2\right)}}\\\dfrac{1}{b+2}\ge\sqrt{\dfrac{ca}{\left(c+2\right)\left(a+2\right)}}\\\dfrac{1}{c+2}\ge\sqrt{\dfrac{ab}{\left(a+2\right)\left(b+2\right)}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\ge\dfrac{abc}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)
\(\Leftrightarrow abc\le1< \dfrac{9}{8}\)
Đề sai !
Giả sử \(a=b=c=1\) thay vào phương trình đầu thì :
\(\dfrac{1}{1+2}+\dfrac{1}{1+2}+\dfrac{1}{1+2}=1\) ( Thỏa mãn )
Nhưng \(1.1.1< \dfrac{1}{8}\) ( vô lí )
\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)
Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bđt Cauchy ta có :
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=3\)
Có BĐT: \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Ta có:
\(VT=\)\(\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)
\(=\dfrac{1+1+c^2}{\left(a^2+b^2+1\right)\left(1+1+c^2\right)}+\dfrac{1+1+a^2}{\left(b^2+c^2+1\right)\left(1+1+a^2\right)}+\dfrac{1+1+b^2}{\left(c^2+a^2+1\right)\left(1+1+b^2\right)}\)
Áp dụng BĐT Bunhiacopski cho mẫu số, ta có:
\(\left(a^2+b^2+c^2\right)\left(1+1+c^2\right)\ge\left(a+b+c\right)^2\)
\(\left(b^2+c^2+1\right)\left(1+1+a^2\right)\ge\left(b+c+a\right)^2\)
\(\left(c^2+a^2+1\right)\left(1+1+b^2\right)\ge\left(c+a+b\right)^2\)
\(\Rightarrow VT\le\dfrac{1+1+c^2}{\left(a+b+c\right)^2}+\dfrac{1+1+a^2}{\left(b+c+a\right)^2}+\dfrac{1+1+b^2}{\left(c+a+b\right)^2}=\dfrac{6+a^2+b^2+c^2}{\left(a+b+c\right)^2}\le\dfrac{6+ab+bc+ca}{3\left(ab+bc+ca\right)}=\dfrac{6+3}{3.3}=1\)
\("="\Leftrightarrow a=b=c=1\)