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Ta có:
\(\sum\dfrac{ab+c}{c+1}=\sum\dfrac{ab+c}{a+c+b+c}\le\sum\dfrac{ab+c}{4}.\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=\dfrac{a+b+c+3}{4}=\dfrac{4}{4}=1\)
Ta có : \(\frac{ab+c}{c+1}=\frac{ab+c\left(a+b+c\right)}{c+a+b+c}=\frac{a\left(b+c\right)+c\left(b+c\right)}{c+a+b+c}=\frac{\left(a+c\right)\left(b+c\right)}{c+a+b+c}\)
Do \(a;b;c>0\Rightarrow a+c;b+c>0\)
Áp dụng BĐT phụ : \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) , ta có :
\(\frac{ab+c}{c+1}\le\frac{\left(a+c\right)\left(b+c\right)}{4}\left(\frac{1}{c+a}+\frac{1}{b+c}\right)=\frac{\left(a+c\right)\left(b+c\right)}{4}.\frac{a+b+c+c}{\left(a+c\right)\left(b+c\right)}=\frac{c+1}{4}\left(1\right)\)
Tương tự , ta có : \(\frac{bc+a}{a+1}\le\frac{a+1}{4}\) ; \(\frac{ac+b}{b+1}\le\frac{b+1}{4}\left(2\right)\)
Từ ( 1 ) ; ( 2 ) có : \(\frac{ab+c}{c+1}+\frac{bc+a}{a+1}+\frac{ac+b}{b+1}\le\frac{a+1+b+1+c+1}{4}=\frac{a+b+c+3}{4}=1\)
Dấu " = " xảy ra <=> \(a=b=c=\frac{1}{3}\)
Vậy ...
Lời giải:
Áp dụng BĐT AM-GM:
\(a^3+a\geq 2a^2; b^3+b\geq 2b^2; c^3+c\geq 2c^2\)
\(\Rightarrow A=\frac{a}{a^3+a+1}+\frac{b}{b^3+b+1}+\frac{c}{c^3+c+1}\leq \frac{a}{2a^2+1}+\frac{b}{2b^2+1}+\frac{c}{2c^2+1}\)
\(\leq \frac{a}{a^2+2a}+\frac{b}{b^2+2b}+\frac{c}{c^2+2c}\)
hay \(A\leq \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}(1)\)
Vì $abc=1$ nên đặt \((a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})(x,y,z>0)\)
Khi đó:
\(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=\frac{y}{x+2y}+\frac{z}{y+2z}+\frac{x}{z+2x}=\frac{1}{2}(1-\frac{x}{x+2y}+1-\frac{y}{y+2z}+1-\frac{z}{z+2x})\)
\(=\frac{3}{2}-\frac{1}{2}(\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2zy}+\frac{z^2}{z^2+2xz})\)
\(\leq \frac{3}{2}-\frac{1}{2}.\frac{(x+y+z)^2}{x^2+2xy+y^2+2zy+z^2+2xz}=\frac{3}{2}-\frac{1}{2}.\frac{(x+y+z)^2}{(x+y+z)^2}=1(2)\) (theo BĐT Cauchy-Schwarz)
Từ \((1);(2)\Rightarrow A\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
bai n ay la bai o dau ma dau cung thay nhi
\(\left(a^{\dfrac{1}{3}};b^{\dfrac{1}{3}};c^{\dfrac{1}{3}}\right)->\left(x;y;z>0\right)\left(xyz=1\right)\)\(\RightarrowΣ\dfrac{x^3}{x^9+x^3+1}\le1\)
\(\dfrac{x^3}{x^9+x^3+1}\le\dfrac{x^2+1}{2\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow-\dfrac{\left(x-1\right)^2\left(x^9+2x^8+4x^7+6x^6+6x^5+6x^4+5x^3+4x^2+2x+1\right)}{2\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^9+x^3+\right)}\le0\)
\(\Rightarrow VT\le\dfrac{1}{2}\cdot2=1=VP\)
a=b=c=x=y=z=1
\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{b\left(c+1+ac\right)}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{1}{b+1+bc}+\dfrac{1}{c+1+ac}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac}{abc+ac+abc.c}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac}{1+ac+c}+\dfrac{1}{ac+c+c}+\dfrac{c}{ac+c+1}\)
\(=\dfrac{ac+1+c}{ac+c+1}=1\) (đpcm)
Ta có \(\dfrac{ab+c}{c+1}=\dfrac{ab+c\left(a+b+c\right)}{\left(a+c\right)+\left(b+c\right)}=\dfrac{\left(a+c\right)\left(b+c\right)}{\left(a+c\right)+\left(b+c\right)}\)
\(\Rightarrow VT=\dfrac{\left(a+c\right)\left(b+c\right)}{\left(a+c\right)+\left(b+c\right)}+\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a+b\right)+\left(b+c\right)}+\dfrac{\left(a+c\right)\left(a+b\right)}{\left(a+b\right)+\left(a+c\right)}\)
Đặt \(\left\{{}\begin{matrix}a+c=x\\b+c=y\\a+b=z\end{matrix}\right.\) \(\Rightarrow x+y+z=2\)
\(\Rightarrow VT\Leftrightarrow\dfrac{xy}{x+y}+\dfrac{yz}{z+y}+\dfrac{xz}{x+z}\)
Áp dụng bất đẳng thức \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{xy}{x+y}\le\dfrac{xy}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{y}{4}+\dfrac{x}{4}\)
Thiết lập tương tự và thu lại ta có
\(\Rightarrow VT\le\dfrac{2\left(x+y+z\right)}{4}=1\) ( đpcm )
\(\Leftrightarrow\dfrac{ab+c}{c+1}+\dfrac{bc+a}{a+1}+\dfrac{ac+b}{b+1}\le1\)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)