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a/ \(\frac{b}{b}.\sqrt{\frac{a^2+b^2}{2}}+\frac{c}{c}.\sqrt{\frac{b^2+c^2}{2}}+\frac{a}{a}.\sqrt{\frac{c^2+a^2}{2}}\)
\(\le\frac{1}{b}.\left(\frac{3b^2+a^2}{4}\right)+\frac{1}{c}.\left(\frac{3c^2+b^2}{4}\right)+\frac{1}{a}.\left(\frac{3a^2+c^2}{4}\right)\)
\(=\frac{1}{4}.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{3}{4}.\left(a+b+c\right)\)
Ta cần chứng minh
\(\frac{1}{4}.\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\frac{3}{4}.\left(a+b+c\right)\le\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
\(\Leftrightarrow\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\left(a+b+c\right)\)
Mà: \(\Leftrightarrow\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\ge\frac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Vậy có ĐPCM.
Câu b làm y chang.
Biến đổi tương đương :
\(\left|a-b\right|+\left|b-c\right|+\left|c-a\right|\ge\sqrt{a^2+b^2+c^2-ab-bc-ac}\)
\(\Leftrightarrow4\left|a-b\right|+4\left|b-c\right|+4\left|c-a\right|\ge\sqrt{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(\Leftrightarrow4\left|a-b\right|+4\left|b-c\right|+4\left|c-a\right|\ge\sqrt{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
Đặt \(\left|a-b\right|=x;\left|b-c\right|=y;\left|c-a\right|=z\)
\(BĐT\Leftrightarrow4x+4y+4z\ge\sqrt{x^2+y^2+z^2}\)
\(\Leftrightarrow16\left(x^2+y^2+z^2+2xy+2yz+2xy\right)\ge x^2+y^2+z^2\)
\(\Leftrightarrow15x^2+15y^2+15z^2+32xy+32yz+32xz\ge0\) (luôn đúng vì \(x;y;z\ge0\))
Vậy \(\left|a-b\right|+\left|b-c\right|+\left|c-a\right|\ge\sqrt{a^2+b^2+c^2-ab-bc-ac}\)
\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{\left(a+\sqrt{bc}\right)^2}=a+\sqrt{bc}\)
Tương tự: \(\sqrt{b+ac}\ge b+\sqrt{ac}\) ; \(\sqrt{c+ab}\ge c+\sqrt{ab}\)
\(\Rightarrow VT\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}\)
\(\Rightarrow VT\ge a+b+c=1\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
1.
Áp dụng BĐT Cauchy - Schwars ta có:
\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}=\sqrt{\left(\sqrt{a}^2+\sqrt{b}^2\right)\left(\sqrt{a}^2+\sqrt{c}^2\right)}\ge a+\sqrt{bc}\).
Tương tự rồi cộng vế với vế ta có đpcm.
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
Biến đổi tương đương:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\)
\(\Leftrightarrow\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\ge\frac{2}{\sqrt{ab}}+\frac{2}{\sqrt{ac}}+\frac{2}{\sqrt{bc}}\)
\(\Leftrightarrow\frac{1}{a}-\frac{2}{\sqrt{ab}}+\frac{1}{b}+\frac{1}{a}-\frac{2}{\sqrt{ac}}+\frac{1}{c}+\frac{1}{b}-\frac{2}{\sqrt{bc}}+\frac{1}{c}\ge0\)
\(\Leftrightarrow\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2+\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{c}}\right)^2+\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{c}}\right)^2\ge0\) (luôn đúng)
Vậy BĐT được chứng minh, dấu "=" xảy ra khi \(a=b=c\)
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 \(x=y=z;x,y,z>0\)
Áp dụng bất đẳng thức \(AM-GM\) cho 2 số dương ta có:
\(\left\{{}\begin{matrix}\dfrac{a+b}{2}\ge\sqrt{ab}\\\dfrac{b+c}{2}\ge\sqrt{bc}\\\dfrac{a+c}{2}\ge\sqrt{ac}\end{matrix}\right.\)
Cộng theo 3 vế ta có:
\(\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{a+c}{2}\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(\Rightarrow\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}b+\dfrac{1}{2}c+\dfrac{1}{2}a+\dfrac{1}{2}c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
\(\Rightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\left(đpcm\right)\)
\(a=b=c\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=2ab\\b^2+c^2=2bc\\a^2+c^2=2ac\end{matrix}\right.\)
Cộng theo 3 vế ta có:
\(a^2+b^2+b^2+c^2+a^2+c^2=2ab+2bc+2ac\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
Ngược lại,khi \(a\ne b\ne c\) thì \(\left\{{}\begin{matrix}a^2+b^2>2ab\\b^2+c^2>2bc\\a^2+c^2>2ac\end{matrix}\right.\) ta có thể dễ dàng cm được \(a^2+b^2+c^2>ab+bc+ac\)