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TỪ GT => \(3\le xy+yz+zx\)
=> \(P\ge\frac{x^3}{\sqrt{y^2+xy+yz+zx}}+\frac{y^3}{\sqrt{z^2+xy+yz+zx}}+\frac{z^3}{\sqrt{x^2+xy+yz+zx}}\)
=> \(P\ge\frac{x^3}{\sqrt{\left(x+y\right)\left(y+z\right)}}+\frac{y^3}{\sqrt{\left(z+x\right)\left(z+y\right)}}+\frac{z^3}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)
TA ÁP DỤNG BĐT CAUCHY 2 SỐ SẼ ĐƯỢC:
=> \(\hept{\begin{cases}\sqrt{x+y}.\sqrt{y+z}\le\frac{x+2y+z}{2}\\\sqrt{z+x}.\sqrt{z+y}\le\frac{x+y+2z}{2}\\\sqrt{x+y}.\sqrt{x+z}\le\frac{2x+y+z}{2}\end{cases}}\)
=> \(P\ge\frac{2x^3}{x+2y+z}+\frac{2y^3}{x+y+2z}+\frac{2z^3}{2x+y+z}\)
=> \(P\ge\frac{2x^4}{x^2+2xy+2xz}+\frac{2y^4}{xy+y^2+2yz}+\frac{2z^4}{2xz+yz+z^2}\)
TA TIẾP TỤC ÁP DỤNG BĐT CAUCHY - SCHWARZ SẼ ĐƯỢC:
=> \(P\ge\frac{2\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)
TA CÓ 1 BĐT SAU: \(xy+yz+zx\le x^2+y^2+z^2\) (*)
=> \(P\ge\frac{2\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(x^2+y^2+z^2\right)}\)
=> \(P\ge\frac{2\left(x^2+y^2+z^2\right)^2}{4\left(x^2+y^2+z^2\right)}=\frac{x^2+y^2+z^2}{2}\)
TA LẠI 1 LẦN NỮA SỬ DỤNG BĐT (*) SẼ ĐƯỢC:
=> \(P\ge\frac{xy+yz+zx}{2}\ge\frac{3}{2}\left(gt\right)\)
DẤU "=" XẢY RA <=> \(x=y=z\)
VẬY P MIN \(=\frac{3}{2}\Leftrightarrow x=y=z=1\)
Ta có :
\(P\ge\frac{x^3}{\sqrt{y^2+xy+yz+zx}}+\frac{y^3}{\sqrt{z^2+xy+yz+zx}}+\frac{z^3}{\sqrt{z^2+xy+yz+zx}}\)
\(=\frac{x^3}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\frac{y^3}{\sqrt{\left(z+x\right)\left(z+y\right)}}+\frac{z^3}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)
\(\ge\frac{2x^3}{x+2y+z}+\frac{2y^3}{x+y+2z}+\frac{2z^3}{2x+y+z}\)\(\ge2.\frac{\left(x^2+y^2+z^2\right)^2}{\left(x^2+y^2+z^2\right)+3.\left(xy+yz+zx\right)}\ge2.\frac{\left(xy+yz+zx\right)^2}{4.\left(xy+yz+zx\right)}\ge2.\frac{3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
ta có:
\(S\ge\frac{x^3}{x^2+y^2+\frac{x^2+y^2}{2}}+\frac{y^3}{y^2+z^2+\frac{y^2+z^2}{2}}+\frac{z^3}{z^2+x^2+\frac{z^2+x^2}{2}}\)
\(\Rightarrow S\ge\frac{2x^3}{3\left(x^2+y^2\right)}+\frac{2y^3}{3\left(y^2+z^2\right)}+\frac{2z^3}{3\left(z^2+x^2\right)}\Rightarrow\frac{3}{2}S\ge P=\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}\)
\(\Rightarrow P=x-\frac{xy^2}{x^2+y^2}+y-\frac{yz^2}{y^2+z^2}+z-\frac{zx^2}{z^2+x^2}\ge\left(x+y+z\right)-\left(\frac{xy^2}{2xy}+\frac{yz^2}{2yz}+\frac{zx^2}{2xz}\right)\)
\(=\left(x+y+z\right)-\frac{1}{2}\left(x+y+z\right)=\frac{9}{2}\)
\(\Rightarrow\frac{3}{2}S\ge\frac{9}{2}\Rightarrow S\ge3\)
Vậy Min S=3 khi x=y=z=3
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Câu 2: \(\left(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\right)^2=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2+2\left(x^2+y^2+z^2\right)\)
\(=\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2+6\)
Áp dụng bất đẳng thức AM - GM ta có :
\(\left(\frac{xy}{z}\right)^2+\left(\frac{yz}{x}\right)^2+\left(\frac{xz}{y}\right)^2\ge3\sqrt[3]{\left(\frac{xy}{z}\right)^2\left(\frac{yz}{x}\right)^2\left(\frac{xy}{y}\right)^2}=3\sqrt[3]{\frac{\left(xyz\right)^4}{\left(xyz\right)^2}}=3\)\(\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge\sqrt{3+6}=3\left(dpcm\right)\)
tại sao lại suy ra đc \(3\sqrt[3]{\frac{\left(xyz\right)^4}{\left(xyz\right)^{^2}}}=3\) vậy cậu?
Vì đã khuya nên não cũng không còn hoạt động tốt nữa, mình làm bài 1 thôi nhé.
Bài 1:
a)
\(2\text{VT}=\sum \frac{2bc}{a^2+2bc}=\sum (1-\frac{a^2}{a^2+2bc})=3-\sum \frac{a^2}{a^2+2bc}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\sum \frac{a^2}{a^2+2bc}\geq \frac{(a+b+c)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{(a+b+c)^2}{(a+b+c)^2}=1\)
Do đó: \(2\text{VT}\leq 3-1\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
b)
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\sum \frac{ab^2}{a^2+2b^2+c^2}=\sum \frac{ab^2}{\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+b^2}\leq \sum \frac{1}{16}\left(\frac{9ab^2}{a^2+b^2+c^2}+\frac{ab^2}{b^2}\right)\)
\(=\frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2}+\frac{a+b+c}{16}(1)\)
Áp dụng BĐT AM-GM:
\(3(ab^2+bc^2+ca^2)\leq (a^2+b^2+c^2)(a+b+c)\)
\(\Rightarrow \frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2)}\leq \frac{3}{16}(a+b+c)(2)\)
Từ $(1);(2)\Rightarrow \text{VT}\leq \frac{a+b+c}{4}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Lý giải xíu chỗ $3(ab^2+bc^2+ca^2)\leq (a^2+b^2+c^2)(a+b+c)$ cho bạn nào chưa rõ:
Áp dụng BĐT AM-GM:
$(a^2+b^2+c^2)(a+b+c)=(a^3+ac^2)+(b^3+a^2b)+(c^3+b^2c)+(ab^2+bc^2+ca^2)$
$\geq 2a^2c+2ab^2+2bc^2+(ab^2+bc^2+ca^2)=3(ab^2+bc^2+ca^2)$
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(\text{VT}=x-\frac{x}{x^2+z}+y-\frac{y}{y^2+x}+z-\frac{z}{z^2+y}=(x+y+z)-\left(\frac{x}{x^2+z}+\frac{y}{y^2+x}+\frac{z}{z^2+y}\right)\)
\(\geq (x+y+z)-\left(\frac{x}{2\sqrt{x^2z}}+\frac{y}{2\sqrt{y^2x}}+\frac{z}{2\sqrt{z^2y}}\right)=(x+y+z)-\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)(1)\)
Từ giả thiết \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Cauchy-Schwarz:
\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3(2)\)
\(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\leq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(1+1+1)=9\)
\(\Rightarrow \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)\leq 3(3)\)
Từ \((1);(2);(3)\Rightarrow \text{VT}\geq 3-\frac{1}{2}.3=\frac{3}{2}\)
Mặt khác: \(\text{VP}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{2}\)
Do đó \(\text{VT}\geq \text{VP}\) (đpcm)
Dấu "=" xảy ra khi $x=y=z=1$