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a) Áp dụng bất đẳng thức AM-GM ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{abc^2}{ab}}=2\sqrt{c^2}=2\left|c\right|=2c\left(c>0\right)\)
Chứng minh tương tự ta được: \(\left\{{}\begin{matrix}\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\\\dfrac{bc}{a}+\dfrac{ab}{c}\ge2b\end{matrix}\right.\)
Cộng theo vế: \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\left(đpcm\right)\)
Áp dụng liên tiếp AM-GM và Cauchy-Schwarz ta được:
\(\dfrac{ab}{a+b}=\dfrac{ab+b^2-b^2}{a+b}=\dfrac{b\left(a+b\right)}{a+b}-\dfrac{b^2}{a+b}=b-\dfrac{b^2}{a+b}\)
Chứng minh tương tự:
\(\left\{{}\begin{matrix}\dfrac{bc}{b+c}=\dfrac{bc+c^2-c^2}{b+c}=\dfrac{c\left(b+c\right)}{b+c}-\dfrac{c^2}{b+c}=c-\dfrac{c^2}{b+c}\\\dfrac{ac}{c+a}=\dfrac{ac+a^2-a^2}{c+a}=\dfrac{a\left(c+a\right)}{c+a}-\dfrac{a^2}{c+a}=a-\dfrac{a^2}{c+a}\end{matrix}\right.\)
Cộng theo vế:
\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ac}{a+c}=a+b+c-\left(\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+c}\right)\le\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\left(đpcm\right)\)
b)Đặt \(A=\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\)
\(A=\dfrac{a\left(a+b\right)-a^2}{a+b}+\dfrac{b\left(b+c\right)-b^2}{a+b}+\dfrac{c\left(c+a\right)-c^2}{c+a}\)
\(A=a+b+c-\dfrac{a^2}{a+b}-\dfrac{b^2}{b+c}-\dfrac{c^2}{c+a}\)
Lại có:\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
\(\Rightarrow A\le a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)
\(\Rightarrowđpcm\)
e)
\(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng)
=> ĐPCM
2a)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
2b)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1)
Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Giải
Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0
Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)
\(\Rightarrow\dfrac{1}{x+2y}+\dfrac{1}{2x+y}\ge\dfrac{4}{x+2y+2x+y}=\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)
Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y
Lời giải:
\(\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c\)
\(\Leftrightarrow \frac{a^2+bc}{b+c}-c+\frac{b^2+ac}{a+c}-a+\frac{c^2+ab}{a+b}-b\geq 0\)
\(\Leftrightarrow \frac{a^2-c^2}{b+c}+\frac{b^2-a^2}{a+c}+\frac{c^2-b^2}{a+b}\geq 0\)
\(\Leftrightarrow a^2\left(\frac{1}{b+c}-\frac{1}{a+c}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\geq 0\)
\(\Leftrightarrow \frac{a^2(a-b)(a+b)+b^2(b-c)(b+c)+c^2(c-a)(c+a)}{(a+b)(b+c)(c+a)}\geq 0\)
\(\Leftrightarrow a^2(a^2-b^2)+b^2(b^2-c^2)+c^2(c^2-a^2)\geq 0\)
\(\Leftrightarrow a^4+b^4+c^4-(a^2b^2+b^2c^2+c^2a^2)\geq 0\)
\(\Leftrightarrow \frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{2}\geq 0\) (luôn đúng)
Do đó ta có đpcm
Dấu bằng xảy ra khi $a=b=c$
Xét hiệu VT - VP
\(\dfrac{a+b}{bc+a^2}+\dfrac{b+c}{ab+b^2}+\dfrac{c+a}{ab+c^2}-\dfrac{1}{a}-\dfrac{1}{b}-\dfrac{1}{c}=\dfrac{a^2+ab-bc-a^2}{a\left(bc+a^2\right)}+\dfrac{b^2+bc-ac-b^2}{b\left(ac+b^2\right)}+\dfrac{c^2+ac-ab-c^2}{c\left(ab+c^2\right)}=\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}\)
Do a,b,c bình đẳng nên giả sử a\(\ge\)b\(\ge\)c, khi đó \(b\left(a-c\right)\)\(\ge\)0, c(b-a)\(\le\)0, a(c-b)\(\le\)0
\(a^3\ge b^3\ge c^3=>abc+a^3\ge abc+b^3\ge abc+c^3\)=>\(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}\le\dfrac{b\left(a-c\right)}{b\left(ac+b^2\right)}\)
=> VT -VP \(\le\) \(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}=\dfrac{ab-ac}{b\left(ac+b^2\right)}+\dfrac{ac-ab}{c\left(ab+c^2\right)}=\dfrac{a\left(b-c\right)}{b\left(ac+b^2\right)}-\dfrac{a\left(b-c\right)}{c\left(ab+c^2\right)}\)
mà \(\dfrac{1}{b\left(ac+b^2\right)}\le\dfrac{1}{c\left(ab+c^2\right)}\) nên VT-VP <0 đpcm
a)Svac-so:
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2\left(đpcm\right)}\)
b)\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\)
\(\Leftrightarrow\dfrac{1}{a^2+1}-\dfrac{1}{ab+1}+\dfrac{1}{b^2+1}-\dfrac{1}{ab+1}\ge0\)
\(\Leftrightarrow\dfrac{ab+1-a^2-1}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{ab+1-b^2-1}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{a\left(b-a\right)}{\left(a^2+1\right)\left(ab+1\right)}+\dfrac{b\left(a-b\right)}{\left(b^2+1\right)\left(ab+1\right)}\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b}{\left(b^2+1\right)\left(ab+1\right)}-\dfrac{a}{\left(a^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{b\left(a^2+1\right)-a\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{a^2b+b-ab^2-a}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{ab\left(a-b\right)-\left(a-b\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\cdot\dfrac{ab-1}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)(luôn đúng)
Áp dụng BĐT Cauchy ta có
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge a\)
\(\dfrac{b^2}{a+c}+\dfrac{a+c}{4}\ge b\)
\(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c
Làm tắt vài chỗ thông cảm
Câu b,
Ta có BĐT Cauchy \(a^2+b^2\ge2ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)
\(\Rightarrow\dfrac{ab}{a+b}\le\dfrac{\left(a+b\right)^2}{4\left(a+b\right)}=\dfrac{a+b}{4}\)
Tương tự \(\dfrac{bc}{b+c}\le\dfrac{b+c}{4}\)
\(\dfrac{ac}{a+c}\le\dfrac{a+c}{4}\)
Cộng theo vế ta đc \(VT\le\dfrac{2\left(a+b+c\right)}{4}=\dfrac{a+b+c}{2}\)
Dấu bằng xảy ra khi a=b=c