\(A=\sqrt{\dfrac{x^2}{x^2+\dfrac{1}{4}xy+y^2}}+\sqrt{\dfrac{y^2}{y^2+\dfrac{1}{4}yz+z^2}}+\sqrt{\dfrac{z^2}{z^2+\dfrac{1}{4}zx+x^2}}\le2\)

\(\Leftrightarrow\sqrt{\dfrac{1}{1+\dfrac{y}{4x}+\dfrac{y^2}{x^2}}}+\sqrt{\dfrac{1}{1+\dfrac{z}{4y}+\dfrac{z^2}{y^2}}}+\sqrt{\dfrac{1}{1+\dfrac{x}{4z}+\dfrac{x^2}{z^2}}}\le2\)

Đặt \(\left\{{}\begin{matrix}\dfrac{y}{x}=a\\\dfrac{z}{y}=b\\\dfrac{x}{z}=c\end{matrix}\right.\) thì bài toán thành

Chứng minh: \(A=\dfrac{1}{\sqrt{4a^2+a+4}}+\dfrac{1}{\sqrt{4b^2+b+4}}+\dfrac{1}{\sqrt{4c^2+c+4}}\le1\) với \(abc=1\)

Thử giải bài toán mới này xem sao bác.

*C/m bài toán mới của HUngnguyen

Ta có BĐT phụ \(\dfrac{1}{\sqrt{4a^2+a+4}}\le\dfrac{a+1}{2\left(a^2+a+1\right)}\)

\(\Leftrightarrow\left(a+1\right)^2\left(4a^2+a+4\right)\ge4\left(a^2+a+1\right)^2\)

\(\Leftrightarrow a\left(a-1\right)^2\ge0\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\dfrac{1}{\sqrt{4b^2+b+4}}\le\dfrac{b+1}{2\left(b^2+b+1\right)};\dfrac{1}{\sqrt{4c^2+c+4}}\le\dfrac{c+1}{2\left(c^2+c+1\right)}\)

CỘng theo vế 3 BĐT trên ta có;

\(VT\le1=VP\) * Chỗ này tự giải chi tiết ra nhé, giờ bận rồi*

Bài này công kềnh vậy thôi thực ra nhìn cái là ra nó là hệ quả của BĐT Vasc của cụ Vasile Bat dang thuc Vasc.pdf

Bài 1:

\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )

Khi đó:

\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)

\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)

\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)

(Áp dụng BĐT Cauchy_Schwarz)

Theo BĐT Cauchy dễ thấy:

\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)

\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$

Bài 2:

Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)

Ta có:

\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)

\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)

Áp dụng BĐT Cauchy:

\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)

\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)

\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)

Nhân theo vế:

\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)

\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)

\(\Rightarrow \text{VT}\leq 1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$

\(\)

Ta có \(2=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Leftrightarrow xy\ge1\)

\(A=\dfrac{1}{x^4+y^2+2xy^2}+\dfrac{1}{x^2+y^4+2x^2y}\\ \le\dfrac{1}{4\sqrt[4]{x^6y^6}}+\dfrac{1}{4\sqrt[4]{x^6y^6}}=\dfrac{1}{4xy}+\dfrac{1}{4xy}\\ \le\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

Dấu \("="\Leftrightarrow x=y=1\)

Chứng minh bằng phép biến đổi tương đương:

1.

\(\Leftrightarrow4+x+y\ge4\sqrt{x+y}\)

\(\Leftrightarrow x+y-4\sqrt{x+y}+4\ge0\)

\(\Leftrightarrow\left(\sqrt{x+y}-2\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

2.

\(\Leftrightarrow\dfrac{y+z}{xyz}\ge\dfrac{4}{x^2+yz}\)

\(\Leftrightarrow\left(y+z\right)\left(x^2+yz\right)\ge4xyz\)

\(\Leftrightarrow x^2y+x^2z+y^2z+z^2y-4xyz\ge0\)

\(\Leftrightarrow y\left(x^2+z^2-2xz\right)+z\left(x^2+y^2-2xy\right)\ge0\)

\(\Leftrightarrow y\left(x-z\right)^2+z\left(x-y\right)^2\ge0\) (đúng)

Ta có :\(\dfrac{x}{y+z}=\dfrac{123-\left(y+z\right)}{y+z}\)

\(\dfrac{y}{x+z}=\dfrac{123-\left(x+z\right)}{x+z}\)

\(\dfrac{z}{y+x}=\dfrac{123-\left(y+x\right)}{y+x}\)

\(\Rightarrow P=\dfrac{123-\left(y+z\right)}{y+z}+\dfrac{123-\left(z+x\right)}{z+x}+\dfrac{123-\left(y+x\right)}{y+x}\)\(\Rightarrow P=123\left(\dfrac{1}{y+z}+\dfrac{1}{x+y}+\dfrac{1}{z+x}\right)-3\)

\(\Rightarrow P=123.\dfrac{1}{45}-3\)

\(\Rightarrow P=-\dfrac{4}{15}\)

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x5=y7=z3=x225=y249=z29" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x5=y7=z3=x225=y249=z29

x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9

=>x=5.9=45

y=7.9=63

z=3*9=27

vậy x=45,y=63,z=27

Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:

\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)

\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)

\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)

\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)

Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)

\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)

\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)

\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)

\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)

\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)

\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)

Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)

c) Đề sai

Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)

Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có 

\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)

\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)

    \(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)