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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
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{a}{\sqrt{1+a^2}}=\dfrac{a}{\sqrt{ab+bc+ca+a^2}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\). Thiết lập 2 BĐT tương tự:
\(\dfrac{b}{\sqrt{1+b^2}}\le\dfrac{1}{4}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{1+c^2}}\le\dfrac{1}{4}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P\le\dfrac{1}{4}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{4}\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Lời giải:
\(\sqrt{a-c}+\sqrt{b-c}=\sqrt{a+b}\)
\(\Leftrightarrow a-c+b-c+2\sqrt{(a-c)(b-c)}=a+b\)
\(\Leftrightarrow c=\sqrt{(a-c)(b-c)}\)
\(\Rightarrow c^2=(a-c)(b-c)=ab-ac-bc+c^2\)
\(\Leftrightarrow ab-ac-bc=0\Leftrightarrow ac+bc-ab=0\)
Ta có: \(P=\frac{bc}{a^2}+\frac{ac}{b^2}-\frac{ab}{c^2}=\frac{(bc)^3+(ac)^3-(ab)^3}{a^2b^2c^2}\) \((1)\)
Ta nhớ đến công thức sau:
\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)\)
(để cm công thức này rất đơn giản bằng việc sử dụng hằng đẳng thức khai triển)
Khi đó, thay thế: \((x,y,z)=(bc,ac,-ab)\)
\(\Rightarrow (bc)^3+(ac)^3-(ab)^3+3a^2b^2c^2=(bc+ac-ab)(.....)=0\)
\(\Rightarrow (bc)^3+(ac)^3-(ab)^3=-3a^2b^2c^2\) (2)
Từ \((1),(2)\Rightarrow P=\frac{-3a^2b^2c^2}{a^2b^2c^2}=-3\)
a.
Xét hiệu:
\(a^3+b^3-ab\left(a+b\right)=\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\)
\(=a^2-ab+b^2-ab=a^2-2ab+b^2\)
\(=\left(a-b\right)^2\ge0\)
=> BĐT luôn đúng
b.
Xét hiệu:
\(a^4+b^4-a^3b-ab^3=\left(a^4-a^3b\right)-\left(b^4-ab^3\right)\)
\(=a^3\left(a-b\right)-b^3\left(a-b\right)=\left(a^3-b^3\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a-b\right)\)
\(=\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
=> BĐT luôn đúng
a)
\(a^3+b^3\ge ab\left(a+b\right)\forall a,b>0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)
\(\Rightarrow a^2-ab+b^2\ge ab\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
\(\Rightarrowđpcm\)
b)
\(a^4+b^4\ge a^3b+ab^3\)
\(\Leftrightarrow a^4-ab^3+b^4-a^3b\ge0\)
\(\Leftrightarrow a\left(a^3-b^3\right)-b\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
\(\Rightarrowđpcm\)
c)
\(\left(a+1\right)\left(b+1\right)\ge\left(\sqrt{ab}+1\right)^2\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)-\left(\sqrt{ab}+1\right)^2\ge0\)
\(\Leftrightarrow1+b+a+ab-ab-2\sqrt{ab}-1\ge0\)
\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)
Dấu bằng xảy ra khi \(a=b\)
d)
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ac\)
Áp dụng bất đẳng thức AM-GM ta được
\(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}.ab}\)
\(\Leftrightarrow\dfrac{a^3}{b}+ab\ge2a^2\)
Tương tự ta được
\(\dfrac{b^3}{c}+bc\ge2b^2,\dfrac{c^3}{a}+ac\ge2c^2\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ac\ge2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ac\right)\)
Mặt khác ta có:\(a^2+b^2+c^2\ge ab+bc+ac\) (hệ quả bất đẳng thức AM-GM)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ac\left(đpcm\right)\)
Dấu bằng xảy ra khi \(x=y=z;x,y,z>0\)
Ta có:\(\dfrac{ab}{a+b}=\dfrac{ab+b^2-b^2}{a+b}=\dfrac{b\left(a+b\right)-b^2}{a+b}=b-\dfrac{b^2}{a+b}\)
Tương tự với các vế ta được:
\(\dfrac{bc}{b+c}=c-\dfrac{c^2}{b+c}\) và \(\dfrac{ac}{a+c}=a-\dfrac{a^2}{a+c}\)
Cộng theo vế:
\(VT=a+b+c-\left(\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+c}\right)\)
Áp dụng bất đẳng thức Cauchy-Schwarz:
\(VT\le a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=a+b+c-\dfrac{a+b+c}{2}=\dfrac{1}{2}\left(a+b+c\right)\)
Ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2c\)
Chứng minh tương tự, ta có:
\(\dfrac{bc}{a}+\dfrac{ab}{c}\ge2b\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\)
\(\Rightarrow2\left(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\right)\ge2\left(a+b+c\right)\)
\(\Rightarrow\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\)
Dấu = xảy ra khi a = b = c
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:\(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
\(\Rightarrow A\le a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\)
\(\Rightarrowđpcm\)
\(A=\sum\sqrt{\dfrac{ab}{c+ab}}=\sum\sqrt{\dfrac{ab}{c^2+ca+cb+ab}}\)
\(=\sum\sqrt{\dfrac{ab}{\left(c+a\right)\left(c+b\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{c+a}+\dfrac{b}{c+b}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{b+a}+\dfrac{c}{b+c}\right)\)
\(=\dfrac{1}{2}.3=\dfrac{3}{2}\)