Ta có:\(\dfrac{x^2}{4}=\dfrac{x}{2};\dfrac{y^2}{9}=\dfrac{y}{3};\dfrac{z^2}{25}=\dfrac{z}{5}\)

Aps dụng tính chất dãy tỉ số bằn nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)

=>\(\dfrac{x}{2}=1=>x=2\)

  \(\dfrac{y}{3}=1=>y=3\)

\(\dfrac{z}{5}=1=>z=5\)

Vậy x=2, y=3, z=5

Ta có : \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được : 

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)

\(\Leftrightarrow x=2;y=3;z=5\)

1.

Ta có:

\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\left(x^2+y^2\right)xy\)

Đặt vế trái của BĐT cần chứng minh là P, áp dụng bồ đề vừa chứng minh ta có:

\(P\le\dfrac{a.abc}{bc\left(b^2+c^2\right)+a.abc}+\dfrac{b.abc}{ca\left(c^2+a^2\right)+b.abc}+\dfrac{c.abc}{ab\left(a^2+b^2\right)+c.abc}\)

\(P\le\dfrac{a^2.bc}{bc\left(a^2+b^2+c^2\right)}+\dfrac{b^2.ac}{ca\left(a^2+b^2+c^2\right)}+\dfrac{c^2.ab}{ab\left(a^2+b^2+c^2\right)}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

2.

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

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

\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)

\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)

\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

\(a,\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)

Lời giải:
Áp dụng BĐT AM-GM:

\(\frac{a^5}{b^2(c+3)}+\frac{b(c+3)}{16}+\frac{ab}{4}\geq \frac{3}{4}a^2\)

Tương tự với các phân thức còn lại và cộng theo vế:

\(A+\frac{5}{16}ab+\frac{3(a+b+c)}{16}\geq \frac{3}{4}(a^2+b^2+c^2)\)

Mà theo BĐT AM-GM dễ thấy \(a^2+b^2+c^2\geq ab+bc+ac\Rightarrow A\geq \frac{7}{16}(a^2+b^2+c^2)-\frac{3}{16}(a+b+c)\)

Áp dụng BĐT AM-GM tiếp:

$a^2+1\geq 2a; b^2+1\geq 2b; c^2+1\geq 2c$

$\Rightarrow a^2+b^2+c^2+3\geq 2(a+b+c)\geq a+b+c+3\sqrt[3]{abc}=a+b+c+3$

$\Rightarrow a^2+b^2+c^2\geq a+b+c\Rightarrow A\geq \frac{1}{4}(a+b+c)\geq \frac{1}{4}\sqrt[3]{abc}=\frac{3}{4}$
Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$

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Cách 2:

Áp dụng BĐT Cauchy-Schwarz:

\(A=\sum \frac{a^6}{ab^2(c+3)}=\sum \frac{a^6}{b+3ab^2}\geq \frac{(a^3+b^3+c^3)^2}{a+b+c+3(ab^2+bc^2+ca^2)}\)$(1)$

Áp dụng BĐT AM-GM:

$a^3+1+1\geq 3a; b^3+1+1\geq 3b; c^3+1+1\geq 3c$

$\Rightarrow a^3+b^3+c^3+6\geq 3(a+b+c)=a+b+c+2(a+b+c)$

$\geq a+b+c+6\sqrt[3]{abc}=a+b+c+6$ (theo BĐT AM-GM)

$\Rightarrow a^3+b^3+c^3\geq a+b+c(2)$

Tiếp tục AM-GM:

$a^3+b^3+b^3\geq 3ab^2; b^3+c^3+c^3\geq 3bc^2; a^3+a^3+c^3\geq 3ca^2$

$\Rightarrow a^3+b^3+c^3\geq ab^2+bc^2+ca^2(3)$

Từ $(1); (2); (3)\Rightarrow A\geq \frac{(a^3+b^3+c^3)^2}{4(a^3+b^3+c^3)}=\frac{a^3+b^3+c^3}{4}\geq \frac{3abc}{4}=\frac{3}{4}$

Ta có đpcm.

\(\dfrac{a^5}{b^2\left(c+3\right)}+\dfrac{b^2}{4}+\dfrac{a\left(c+3\right)}{16}\ge3\sqrt[3]{\dfrac{a^6b^2\left(c+3\right)}{64b^2\left(c+3\right)}}=\dfrac{3}{4}a^2\)

Tương tự: \(\dfrac{b^5}{c^2\left(a+3\right)}+\dfrac{c^2}{4}+\dfrac{b\left(a+3\right)}{16}\ge\dfrac{3}{4}b^2\)

\(\dfrac{c^5}{a^2\left(b+3\right)}+\dfrac{a^2}{4}+\dfrac{c\left(b+3\right)}{16}\ge\dfrac{3}{4}c^2\)

Cộng vế:

\(A+\dfrac{a^2+b^2+c^4}{4}+\dfrac{ab+bc+ca}{16}+\dfrac{9}{16}\ge\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)

\(\Rightarrow A\ge\dfrac{1}{2}\left(a^2+b^2+c^2\right)-\dfrac{ab+bc+ca}{16}-\dfrac{9}{16}\ge\dfrac{1}{2}\left(a^2+b^2+c^2\right)-\dfrac{a^2+b^2+c^2}{16}-\dfrac{9}{16}\)

\(\Rightarrow A\ge\dfrac{7}{16}\left(a^2+b^2+c^2\right)-\dfrac{9}{16}\ge\dfrac{7}{16}.3\sqrt[3]{\left(abc\right)^2}-\dfrac{9}{16}=\dfrac{3}{4}\) (đpcm)