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\(\left(b^3+c^3\right)\left(1+1\right)\left(1+1\right)\ge\left(b+c\right)^3\)
\(\Rightarrow b^3+c^3\ge\dfrac{\left(b+c\right)^3}{4}\Rightarrow\dfrac{a}{\sqrt[3]{b^3+c^3}}\le\dfrac{a\sqrt[3]{4}}{b+c}\)
Tương tự và cộng lại:
\(VT\le\sqrt[3]{4}\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)< \sqrt[3]{4}\left(\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}\right)=2\sqrt[3]{4}\)
1. Đặt $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=T$
$\frac{a}{b+c}> \frac{a}{a+b+c}$
$\frac{b}{c+a}> \frac{b}{c+a+b}$
$\frac{c}{a+b}> \frac{c}{a+b+c}$
$\Rightarrow T> \frac{a+b+c}{a+b+c}=1$ (đpcm)
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Xét hiệu:
$\frac{a}{b+c}-\frac{2a}{a+b+c}=\frac{-a(b+c-a)}{(b+c)(a+b+c)}<0$ theo BĐT tam giác
$\Rightarrow \frac{a}{b+c}< \frac{2a}{a+b+c}$
Tương tư: $\frac{b}{c+a}< \frac{2b}{c+a+b}$
$\frac{c}{a+b}< \frac{2c}{a+b+c}$
Cộng theo vế:
$T< \frac{2(a+b+c)}{a+b+c}=2$
$\frac{b}{a+c}
2.
Áp dụng BĐT AM-GM:
\(\frac{b+c}{a}.1\leq \frac{1}{4}(\frac{b+c}{a}+1)^2=\frac{(b+c+a)^2}{4a^2}\)
\(\Rightarrow \sqrt{\frac{a}{b+c}}\geq \frac{2a}{a+b+c}\)
Tương tự với các phân thức còn lại và cộng theo vế:
$\Rightarrow T\geq \frac{2(a+b+c)}{a+b+c}=2$
Dấu "=" xảy ra khi $b+c=a; c+a=b; a+b=c\Rightarrow a=b=c=0$ (vô lý)
Vậy dấu "=" không xảy ra, tức là $T>2>1$ (đpcm)
set \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\)\(\Rightarrow x+y+z=3\)
\(VT=\sum\sqrt{\dfrac{\left(x+y\right)\left(x+z\right)}{4x}}=\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}.\left(\sum\dfrac{1}{\sqrt{4x\left(y+z\right)}}\right)\)
Áp dụng BĐT AM-GM:
\(\dfrac{1}{\sqrt{4x\left(y+z\right)}}+\dfrac{1}{\sqrt{4y\left(x+z\right)}}+\dfrac{1}{\sqrt{4z\left(x+y\right)}}\ge\dfrac{9}{2\left(\sqrt{xy+xz}+\sqrt{yz+yx}+\sqrt{xz+zy}\right)}\)
Áp dụng BĐT bunyakovsky:
\(\sum\sqrt{xy+yz}\le\sqrt{6\left(xy+yz+xz\right)}\)
\(\Rightarrow\sum\dfrac{1}{2\sqrt{x\left(y+z\right)}}\ge\dfrac{9}{2\sqrt{6\left(xy+yz+xz\right)}}\)
Mà \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{3}\left(xy+yz+xz\right)\)(*)
\(\Rightarrow VT\ge\sqrt{\dfrac{8}{3}\left(xy+yz+xz\right)}.\dfrac{9}{2\sqrt{6\left(xy+yz+xz\right)}}=3\)
Dấu = xảy ra khi x=y=z hay a=b=c=1
(*) Prove BĐT \(\left(m+n\right)\left(n+p\right)\left(m+p\right)\ge\dfrac{8}{9}\left(m+n+p\right)\left(mn+np+pm\right)\)
khai triển ,để ý rằng \(\left(m+n\right)\left(n+p\right)\left(p+m\right)=\left(m+n+p\right)\left(mn+np+pm\right)-mnp\)
Đặt \(\left(a;b;c\right)=\left(x^4;y^4;z^4\right)\Rightarrow xyz=1\)
\(VT=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)
\(VT=\dfrac{1}{x^2+y^2+y^2+1+2}+\dfrac{1}{y^2+z^2+z^2+1+2}+\dfrac{1}{z^2+x^2+x^2+1+2}\)
\(VT\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có:
\(\left(2a^2-b^2-c^2\right)^2\ge0\)
\(\Leftrightarrow4a^4+b^4+c^4-4a^2b^2-4a^2c^2+2b^2c^2\ge0\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\ge6a^2b^2+6a^2c^2-3a^4\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge3a^2\left(2b^2+2c^2-a^2\right)\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2b^2+2c^2-a^2}}\ge\dfrac{\sqrt{3}a}{a^2+b^2+c^2}\)
\(\Leftrightarrow\dfrac{a}{\sqrt{2b^2+2c^2-a^2}}\ge\sqrt{3}\dfrac{a^2}{a^2+b^2+c^2}\)
Tương tự: \(\dfrac{b}{\sqrt{2a^2+2c^2-b^2}}\ge\sqrt{3}.\dfrac{b^2}{a^2+b^2+c^2}\) ; \(\dfrac{c}{\sqrt{2a^2+2b^2-c^2}}\ge\sqrt{3}.\dfrac{c^2}{a^2+b^2+c^2}\)
Cộng vế: \(P\ge\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=\sqrt{3}\)
\(P_{min}=\sqrt{3}\) khi \(a=b=c\)
Lời giải:
Đặt \(\left ( \frac{\sqrt{a^2+b^2}}{c},\frac{\sqrt{b^2+c^2}}{a}, \frac{\sqrt{c^2+a^2}}{b} \right )=(x,y,z)\)
BĐT cần chứng minh tương đương với:
\(x+y+z\geq 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)\((*)\)
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Từ cách đặt $x,y,z$ ta có:
\(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}=1\)
Áp dụng BĐT Bunhiacopxky:
\(\frac{x^2+1}{x^2}+\frac{y^2+1}{y^2}+\frac{z^2+1}{z^2}=\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\right)\left(\frac{x^2+1}{x^2}+\frac{y^2+1}{y^2}+\frac{z^2+1}{z^2}\right)\)
\(\geq \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
\(\Leftrightarrow 3\geq 2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
\(\Leftrightarrow xyz\geq \frac{2}{3}(x+y+z)\)
\(\Rightarrow xyz(x+y+z)\geq \frac{2}{3}(x+y+z)^2\)
Áp dụng BĐT AM_GM ta lại có:
\((x+y+z)^2\geq 3(xy+yz+xz)\). Do đó:
\(xyz(x+y+z)\geq 2(xy+yz+xz)\)
\(\Leftrightarrow x+y+z\geq 2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Đúng theo \((*)\)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c\)
áp dụng bat dang thuc bunhiacóki
ta có \(\dfrac{\sqrt{a^2+b^2}}{c}\ge\dfrac{a+b}{\sqrt{2}c}\)
ttu vt \(\ge\dfrac{1}{\sqrt{2}}\left(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\right)\)
=\(\dfrac{a}{\sqrt{2}}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{b}{\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+\dfrac{c}{\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) (1)
áp dung bdt \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
ta có (1) \(\ge\dfrac{a}{\sqrt{2}}.\dfrac{4}{b+c}\)
tiếp tục áp dụng bunhia ta có \(\dfrac{a}{\sqrt{2}}.\dfrac{4}{b+c}\ge\dfrac{a}{\sqrt{2}}.\dfrac{4}{\sqrt{2\left(b^2+c^2\right)}}=\dfrac{2a}{\sqrt{b^2+c^2}}\)
ttuong tu ta có \(vt\ge2\left(\dfrac{a}{\sqrt{b^2+c2}}+\dfrac{b}{\sqrt{a^2+c^2}}+\dfrac{c}{\sqrt{a^2+b^2}}\right)\left(dpcm\right)\)
\(\sqrt{\dfrac{a}{b+c-ta}}=\dfrac{a\sqrt{t+1}}{\sqrt{\left(at+a\right)\left(b+c-ta\right)}}\ge\dfrac{2a\sqrt{t+1}}{at+a+b+c-ta}=\dfrac{2a\sqrt{t+1}}{a+b+c}\)
Làm tương tự, cộng lại và rút gọn